7 Differentiation — revision question pack

6 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

7.1 · Derivative of f(x) as tangent gradient and as a limit; rate of change; sketch the gradient function; first-principles differentiation for small integer powers of x, sin x, cos x; second derivatives; convexity, concavity, inflection.

Explanation

The derivative $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ is the tangent gradient and instantaneous rate of change. For $\sin x$ and $\cos x$ , expand the compound angle and use $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$ .
To sketch $f'$ , record where $f$ rises or falls, where its tangents are horizontal, and how steep those tangents are; the zeros of $f'$ occur at stationary points of $f$ . The second derivative is the rate of change of the gradient: $f''(x)>0$ indicates a convex section and $f''(x)<0$ a concave section.
At an inflection point, $f''$ changes sign.
A common error is to declare an inflection point from $f''(x)=0$ alone; a sign change of concavity must be checked.
For first principles, use $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ ; expanding $(x+h)^2$ or $(x+h)^3$ and cancelling before taking the limit derives the required power result.

The derivative at a point is the gradient of the tangent to the curve there.

Worked example

For $f(x)=x^3-3x$ , sketch the gradient function $y=f'(x)$ , marking its intercepts and turning point. Hence state where the graph of $f$ is convex and concave.

1.Differentiate to obtain $f'(x)=3x^2-3=3(x-1)(x+1)$ .
2.Its graph is an upward parabola crossing the $x$ -axis at $x=-1,1$ , with minimum $(0,-3)$ .
3.Since $f''(x)=6x$ , the original graph is concave for $x<0$ and convex for $x>0$ .

Answer: The upward parabola $y=3x^2-3$ , with zeros $(-1,0)$ and $(1,0)$ and minimum $(0,-3)$ ; $f$ is concave for $x<0$ and convex for $x>0$

Common mistakes

Don't sketch $f'$ with zeros at the roots of $f$ rather than at stationary points of $f$ .
Don't confuse the sign of the first derivative with convexity, which is determined by the second derivative.
Don't cancel $h$ but never state the limiting step $h\to0$ in a first-principles proof.

Exam tip

For first principles, state the limiting process explicitly at least once before stating the derivative — $h\to0$ or $\delta x\to0$ , whichever letter you used. On a gradient-function sketch, use zeros for stationary points and whether the gradient is rising or falling for convexity.

Tier 1 · Easy

1.
For $f(x)=x^2-3x$ , use the limit definition of the derivative to find $f'(x)$ .
(4)
(Total for Question 1 is 4 marks)
2.
For $f(x)=x^4-4x^2$ , find $f''(x)$ and state whether the curve is locally convex or concave at $x=1$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Prove, from first principles, that $f'(a)=2a+3$ for $f(x)=x^2+3x$ .
(5)
(Total for Question 1 is 5 marks)
2.
Use first principles to prove that the derivative of $f(x)=x^3+2x$ is $f'(x)=3x^2+2$ .
(5)
(Total for Question 2 is 5 marks)
3.
A cube has edge length $x$ cm and volume $V=x^3$ cm $^3$ . Use first principles to find $\dfrac{\mathrm dV}{\mathrm dx}$ at $x=2$ . Hence estimate the increase in volume when the edge length increases from $2$ cm to $2.03$ cm.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Using first principles and compound-angle identities, prove that $\frac{\mathrm d}{\mathrm dx}(\sin x)=\cos x$ and $\frac{\mathrm d}{\mathrm dx}(\cos x)=-\sin x$ . You may use $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$ .
(8)
(Total for Question 1 is 8 marks)
2.
A differentiable function satisfies $f'(x)=x^2(x-3)$ . Determine the intervals on which $f$ is increasing and decreasing, and classify both stationary points of $f$ .
(6)
(Total for Question 2 is 6 marks)
3.
Let $f(x)=x^3+cx$ , where $c$ is a constant. Use first principles to show that $f'(x)=3x^2+c$ . The tangent to the curve at $x=2$ is perpendicular to the line $y=-\dfrac17x+4$ . Find $c$ and the equation of the tangent.
(6)
(Total for Question 3 is 6 marks)
4.
For a real constant $p$ , let $C$ be the curve $y=x^2+px+7$ . Starting from the limit definition of the derivative, show that the gradient of $C$ at $x=a$ is $2a+p$ . The tangent at $x=2$ passes through $(4,-5)$ . Find $p$ and hence find the vertex of $C$ . Use your limit result to prove that the tangent at the vertex is horizontal, and give its equation.
(7)
(Total for Question 4 is 7 marks)
5.
For the family of curves $f(x)=x^4-4kx^3+6x^2$ , where $k$ is real, determine the values of $k$ for which the curve has exactly two distinct points of inflection. For $k=\dfrac54$ , find both points of inflection and state the intervals on which the curve is convex and concave.
(7)
(Total for Question 5 is 7 marks)

7.2 · Differentiate xⁿ for rational n, and related sums, differences and constant multiples; differentiate e^(kx), a^(kx), sin kx, cos kx, tan kx; understand and use the derivative of ln x.

Explanation

For rational $n$ , $\frac{\mathrm d}{\mathrm dx}(x^n)=nx^{n-1}$ wherever the original expression and derivative are defined.
The standard exponential results are $\frac{\mathrm d}{\mathrm dx}(e^{kx})=ke^{kx}$ and $\frac{\mathrm d}{\mathrm dx}(a^{kx})=k(\ln a)a^{kx}$ .
For angles in radians, $\frac{\mathrm d}{\mathrm dx}(\sin kx)=k\cos kx$ , $\frac{\mathrm d}{\mathrm dx}(\cos kx)=-k\sin kx$ , $\frac{\mathrm d}{\mathrm dx}(\tan kx)=k\sec^2kx$ , and $\frac{\mathrm d}{\mathrm dx}(\ln x)=\frac1x$ .
A common error is to omit the factor $k$ created by the inner function, or the factor $\ln a$ when differentiating $a^{kx}$ .
Differentiate each term and include each inner derivative: $4(3e^{3x})+5(2\cos(2x))-3(1/x)$ .

Worked example

Differentiate $y=4e^{3x}+5\sin(2x)-3\ln x$ .

1.Differentiate each term and include each inner derivative: $4(3e^{3x})+5(2\cos(2x))-3(1/x)$ .
2.Therefore $\frac{\mathrm dy}{\mathrm dx}=12e^{3x}+10\cos(2x)-\frac3x$ .

Answer: $\frac{\mathrm dy}{\mathrm dx}=12e^{3x}+10\cos(2x)-\frac3x$

Common mistakes

Don't differentiate $\ln x$ as $1/\ln x$ instead of $1/x$ .
Don't omit the inner derivative when differentiating exponential or trigonometric functions of a multiple of x.

Exam tip

Differentiate each term separately and display the chain-rule multiplier for every composite term.

Tier 1 · Easy

1.
Differentiate $y=5x^{3/2}-2x^{-1/2}$ with respect to $x$ .
(3)
(Total for Question 1 is 3 marks)
2.
Differentiate $y=5e^{-3x}+2\ln x$ , where $x>0$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Differentiate $y=3x^{5/2}-4x^{-1/2}+2^x$ for $x>0$ .
(4)
(Total for Question 1 is 4 marks)
2.
Differentiate $f(x)=5^{3x}-2\sin(4x)+x^{-3/2}$ for $x>0$ .
(4)
(Total for Question 2 is 4 marks)
3.
The curve $y=2x^{3/2}+e^{-x}+\ln x$ is defined for $x>0$ . Determine the tangent at $x=1$ , giving its equation in exact form.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Given $f(x)=3^{2x}+\tan(5x)-2\cos(3x)$ , find $f'(x)$ and hence find the exact value of $f'(0)$ .
(5)
(Total for Question 1 is 5 marks)
2.
Given $f(x)=2^{3x}+4\sin(2x)-\ln x$ for $x>0$ , find $f'(x)$ and hence find the exact value of $f'(\pi/4)$ .
(5)
(Total for Question 2 is 5 marks)
3.
For $x>0$ , $f(x)=Ax^{3/2}+B\ln x+\sin\!\left(\dfrac{\pi x}{2}\right)$ . Given $f'(1)=6$ and $f'(4)=\dfrac{27}{4}+\dfrac\pi2$ , find $A$ and $B$ . Hence find the exact value of $f'(9)$ .
(6)
(Total for Question 3 is 6 marks)
4.
The function $f$ is given by $f(x)=2^x+a\sin(\pi x)$ , where $a$ is a constant. Given that $f$ has a stationary point at $x=0$ , find $a$ exactly. Hence write its tangent at $x=1$ in exact form.
(6)
(Total for Question 4 is 6 marks)
5.
For $0\leq x\leq2\pi$ , the curve has equation $y=\sin(2x)+2\sin x$ . Find the exact coordinates of every stationary point, showing that your list is complete.
(6)
(Total for Question 5 is 6 marks)

7.3 · Apply differentiation to find gradients, tangents and normals, maxima and minima and stationary points, points of inflection; identify where functions are increasing or decreasing.

Explanation

At $x=a$ , the tangent gradient is $f'(a)$ and a non-vertical normal has gradient $-1/f'(a)$ ; use the point on the curve in point-gradient form. Stationary points solve $f'(x)=0$ .
Classify them by a sign change in $f'$ or, when decisive, by $f''(x)>0$ for a minimum and $f''(x)<0$ for a maximum.
A function is increasing where $f'(x)>0$ and decreasing where $f'(x)<0$ .
For a constrained optimisation problem, include endpoints or domain restrictions in the comparison.
A common error is to assume every solution of $f'(x)=0$ is a maximum or minimum; a stationary point can instead be an inflection point.

Worked example

For $f(x)=x^3-6x^2+9x+2$ , find and classify every stationary point. State the intervals on which $f$ is increasing.

1. $f'(x)=3x^2-12x+9=3(x-1)(x-3)$ , so the stationary values are $x=1,3$ .
2.The coordinates are $f(1)=6$ and $f(3)=2$ .
3.Also $f''(x)=6x-12$ , so $f''(1)=-6$ gives a local maximum and $f''(3)=6$ gives a local minimum.
4.The factorised derivative is positive outside the roots, so $f$ is increasing for $x<1$ and $x>3$ .

Answer: Local maximum $(1,6)$ and local minimum $(3,2)$ ; Increasing for $x<1$ and $x>3$

Common mistakes

Don't use $(x,f'(x))$ as the coordinates of a stationary point instead of $(x,f(x))$ .
Don't find stationary x-values and fail to classify them or test the requested increasing intervals.

Exam tip

For stationary-point questions, solve the derivative equation, classify each point and finish with a derivative sign chart.

Tier 1 · Easy

1.
The curve $y=x^2+3x$ is considered at the point where $x=1$ . Find the equations of the tangent and the normal.
(4)
(Total for Question 1 is 4 marks)
2.
At the point where $x=2$ , determine the tangent to the curve $y=x^3-12x$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find equations of the tangent and the normal to $y=x^2+\dfrac{2}{x}$ at the point where $x=2$ .
(5)
(Total for Question 1 is 5 marks)
2.
Find and classify all stationary points of $f(x)=x^3-3x^2-9x+5$ .
(5)
(Total for Question 2 is 5 marks)
3.
Find the coordinates of the points on the curve $y=x^3-3x$ at which the tangent is parallel to the line $y=9x+2$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A model for the volume of an open container is $V=x(12-x)^2$ for $0<x<12$ , where $V$ is measured in $\text{cm}^3$ . Use calculus to find the maximum possible volume.
(7)
(Total for Question 1 is 7 marks)
2.
A point $P$ lies on the curve $y=\dfrac{12}{x}$ , where $x>0$ . Use calculus to find the exact coordinates of the point on the curve closest to the origin, and find this minimum distance.
(6)
(Total for Question 2 is 6 marks)
3.
A rectangle has vertices $(-x,0)$ , $(x,0)$ , $(-x,12-x^2)$ and $(x,12-x^2)$ , where $0<x<2\sqrt3$ . Use calculus to find the dimensions and maximum area of the rectangle.
(6)
(Total for Question 3 is 6 marks)
4.
For the family of curves $y=x^3-3kx+2$ , where $k$ is real, determine the number and nature of the stationary points for $k<0$ , $k=0$ and $k>0$ . Hence find and classify the stationary points when $k=4$ .
(7)
(Total for Question 4 is 7 marks)
5.
Two distinct tangents to the parabola $y=x^2$ pass through the point $P(0,-4)$ . Find the coordinates of both points of contact and the equations of the two tangents. Hence find an exact expression for the acute angle between the tangents.
(6)
(Total for Question 5 is 6 marks)

7.4 · Differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions.

Explanation

Use $(uv)'=u'v+uv'$ for products and $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$ for quotients; brackets help preserve the order in the quotient numerator. For a composite function, differentiate the outer function and multiply by the inner derivative.
Also $\frac{\mathrm d}{\mathrm dx}(\sec x)=\sec x\tan x$, $\frac{\mathrm d}{\mathrm dx}(\cosec x)=-\cosec x\cot x$ and $\frac{\mathrm d}{\mathrm dx}(\cot x)=-\cosec^2x$ .
Connected rates use a shared variable, for example $\frac{\mathrm dV}{\mathrm dt}=\frac{\mathrm dV}{\mathrm dr}\frac{\mathrm dr}{\mathrm dt}$ .
For an inverse $g=f^{-1}$ , $g'(y)=1/f'(x)$ when $y=f(x)$ and $f'(x)\ne0$ .
A common error is to substitute numerical values before differentiating, which can erase the changing relationship between the variables.

Worked example

(a) Differentiate $y=\frac{\sin(2x)}{(x^2+1)^3}$ , giving one fraction. (b) Differentiate $z=\sec(3x)-2\cot x+\cosec(2x)$ .

1.(a) Take $u=\sin(2x)$ and $v=(x^2+1)^3$ .
2.Then $u'=2\cos(2x)$ and $v'=6x(x^2+1)^2$ .
3.The quotient rule and cancellation of $(x^2+1)^2$ give the stated fraction. (b) Apply the three standard derivatives and the chain rule: the terms give $3\sec(3x)\tan(3x)$ , $2\cosec^2x$ and $-2\cosec(2x)\cot(2x)$ respectively.

Answer: (a) $\frac{\mathrm dy}{\mathrm dx}=\frac{2(x^2+1)\cos(2x)-6x\sin(2x)}{(x^2+1)^4}$ ; (b) $\frac{\mathrm dz}{\mathrm dx}=3\sec(3x)\tan(3x)+2\cosec^2x-2\cosec(2x)\cot(2x)$

Common mistakes

Don't differentiate a product $uv$ as $u'v'$ and omit both product-rule terms.
Don't apply product, quotient and chain rules as separate fragments and lose factors in the combined derivative.

Exam tip

Name the outermost rule first, differentiate nested functions carefully, then simplify only after every factor is present.

Tier 1 · Easy

1.
Differentiate $y=x^2e^{3x}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Differentiate $y=(3x^2-1)^4$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The volume of a sphere is increasing at $12\pi\,\text{cm}^3\text{s}^{-1}$ . Find the rate at which its radius is increasing when the radius is $3\,\text{cm}$ .
(4)
(Total for Question 1 is 4 marks)
2.
Differentiate $y=\dfrac{\ln(2x+1)}{x^2+4}$ , giving your answer as a single fraction.
(4)
(Total for Question 2 is 4 marks)
3.
A point $P(x,y)$ moves along the curve $x^2+4y^2=100$ in the first quadrant. At the instant when $P=(6,4)$ , the $x$ -coordinate is increasing at $2$ units per second. Find the rate of change of the $y$ -coordinate and state whether $P$ is moving upwards or downwards.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
(a) A sphere has radius $r$ cm and volume $V=\frac43\pi r^3$ . At an instant when $r=3$ , its volume is increasing at $24\pi\,\text{cm}^3\text{s}^{-1}$ . Find $\frac{\mathrm dr}{\mathrm dt}$ . (b) The function $f(x)=x^3+x+1$ has inverse $g$ . Given that $g(3)=1$ , find $g'(3)$ .
(7)
(Total for Question 1 is 7 marks)
2.
A conical pile has radius $r$ cm and height $h$ cm, with $h=2r$ at all times. Its volume is $V=\dfrac13\pi r^2h$ . Material is added at $18\pi\,\text{cm}^3\text{s}^{-1}$ . Find $\dfrac{\mathrm dr}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ at the instant when $h=4$ cm.
(6)
(Total for Question 2 is 6 marks)
3.
A rectangle has width $w$ cm and length $l$ cm, where $l=2w+3$ at all times. Its area is increasing at $95\,\text{cm}^2\text{s}^{-1}$ . Find the rates of change of $w$ , $l$ and the perimeter when $w=4$ cm.
(6)
(Total for Question 3 is 6 marks)
4.
A straight ladder of length $10$ m rests with its foot on horizontal ground and its top against a vertical wall. The foot moves away from the wall at $0.6$ m s $^{-1}$ . Let $\theta$ be the angle between the ladder and the ground. Find $\dfrac{\mathrm d\theta}{\mathrm dt}$ and the vertical velocity of the top when the foot is $6$ m from the wall.
(6)
(Total for Question 4 is 6 marks)
5.
For $x>0$ , the curve has equation $y=(x^2+1)^3e^{-x}$ . Differentiate the function in factorised form. Hence find the exact $x$ -coordinates of all stationary points and classify each one.
(6)
(Total for Question 5 is 6 marks)

7.5 · Differentiate simple functions and relations defined implicitly or parametrically, for first derivative only.

Explanation

For an implicit relation, differentiate every term with respect to $x$ and attach a factor $\frac{\mathrm dy}{\mathrm dx}$ whenever a differentiated term contains $y$ .
For parametric equations $x=x(t)$ and $y=y(t)$ , use $\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$ where $\mathrm dx/\mathrm dt\ne0$ .
After finding the gradient, use the parameter or relation to obtain the actual point before writing a tangent or normal equation.
A common error in implicit differentiation is to write $\frac{\mathrm d}{\mathrm dx}(xy)=y+x$ instead of $y+x\frac{\mathrm dy}{\mathrm dx}$ .

Worked example

A curve has parametric equations $x=t^2+1$ and $y=t^3-3t$ . Find the equation of its tangent when $t=2$ .

1. $\frac{\mathrm dx}{\mathrm dt}=2t$ and $\frac{\mathrm dy}{\mathrm dt}=3t^2-3$ , so $\frac{\mathrm dy}{\mathrm dx}=\frac{3t^2-3}{2t}$ .
2.At $t=2$ , the gradient is $\frac94$ .
3.The point is $(5,2)$ , so the tangent is $y-2=\frac94(x-5)$ .

Answer: $y-2=\frac94(x-5)$

Common mistakes

Don't substitute the parameter before forming $\mathrm dy/\mathrm dx$ , concealing a zero value of $\mathrm dx/\mathrm dt$ .
Don't calculate the two parametric derivatives but use their product instead of dy/dx equals (dy/dt)/(dx/dt).

Exam tip

For a parametric tangent, evaluate both parameter derivatives at the stated parameter before forming the gradient and line equation.

Tier 1 · Easy

1.
The curve obeys $x^2+y^2=25$ . By implicit differentiation, obtain its gradient at a general point $(x,y)$ .
(3)
(Total for Question 1 is 3 marks)
2.
A curve has parametric equations $x=t^2+1$ and $y=5t-2$ . Find $\dfrac{\mathrm dy}{\mathrm dx}$ in terms of $t$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The curve $x^2+xy+y^2=7$ passes through $(1,2)$ . Obtain a formula for its gradient, then determine the tangent at $(1,2)$ .
(5)
(Total for Question 1 is 5 marks)
2.
A curve has parametric equations $x=t^3+1$ and $y=t^2-4t$ . Find the value of $t$ at which the curve has a horizontal tangent, and find the equations of the tangent and normal there.
(5)
(Total for Question 2 is 5 marks)
3.
A curve has parametric equations $x=2\cos t$ and $y=3\sin t$ . Find the exact equation of the tangent when $t=\dfrac\pi4$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The curve has parametric equations $x=t+\frac1t$ and $y=t-\frac1t$ , where $t>0$ . Find the exact equations of the tangent and normal at $t=2$ .
(7)
(Total for Question 1 is 7 marks)
2.
The curve $x^2+xy+2y^2=8$ passes through $P(2,1)$ . The normal at $P$ meets the coordinate axes at $A$ and $B$ . Find the exact area of triangle $OAB$ , where $O$ is the origin.
(7)
(Total for Question 2 is 7 marks)
3.
A curve has parametric equations $x=t^2+t$ and $y=t^3$ . Find all values of $t$ for which the tangent has gradient $1$ , and find the equation of each tangent.
(6)
(Total for Question 3 is 6 marks)
4.
A curve has parametric equations $x=t^2-1$ and $y=t^3-t$ . The curve passes through the origin for two distinct values of $t$ . Find these values and the equation of the tangent corresponding to each value. Show that the two tangents are perpendicular.
(6)
(Total for Question 4 is 6 marks)
5.
The curve $x^2+2xy+3y^2=12$ has both horizontal and vertical tangents. By implicit differentiation, find the exact coordinates of every point with a horizontal tangent and every point with a vertical tangent.
(7)
(Total for Question 5 is 7 marks)

7.6 · Construct simple differential equations in pure mathematics and in context (contexts may include kinematics, population growth and modelling the relationship between price and demand).

Explanation

Translate a rate statement into derivative notation after defining the dependent and independent variables, including their units where relevant.
Phrases such as 'proportional to' introduce a positive constant $k$ ; words such as 'decreases' or 'decays' determine whether a minus sign is needed.
A limiting or equilibrium value often appears as a difference, for example growth towards a capacity $K$ can be modelled by $\frac{\mathrm dP}{\mathrm dt}=k(K-P)$ .
A common error is to solve the differential equation when only its construction is requested, while failing to state or determine the proportionality constant.

Worked example

A population $P$ grows at a rate proportional to the difference between $1200$ and the current population. When $P=800$ , the population is increasing at $50$ individuals per year. Construct the differential equation, including the value of the constant of proportionality.

1.Write $\frac{\mathrm dP}{\mathrm dt}=k(1200-P)$ .
2.Using $P=800$ and rate $50$ gives $50=k(400)$ , so $k=\frac18$ .
3.Therefore $\frac{\mathrm dP}{\mathrm dt}=\frac18(1200-P)$ .

Answer: $\frac{\mathrm dP}{\mathrm dt}=\frac18(1200-P)$

Common mistakes

Don't assign the wrong sign to a decay or resistance term, predicting growth when the quantity should fall.
Don't write a proportional relationship without defining the constant or using the supplied rate to determine it.

Exam tip

Translate each verbal rate statement into a signed differential equation, then substitute the calibration data to find the constant.

Tier 1 · Easy

1.
The acceleration of a particle is proportional to its speed $v$ and acts opposite to the motion. Write down a differential equation for $v$ in terms of time $t$ .
(2)
(Total for Question 1 is 2 marks)
2.
The rate of change of $y$ with respect to $x$ is proportional to $x^2$ . Construct a differential equation relating $y$ and $x$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A population $P$ grows at a rate proportional to the product $P(1000-P)$ . When $P=200$ , the population is increasing at $32$ individuals per day. Construct the differential equation, including the constant of proportionality.
(4)
(Total for Question 1 is 4 marks)
2.
A particle moves in a straight line with speed $v$ . Its acceleration is $6$ minus a quantity proportional to $v^2$ . When $v=2$ , its acceleration is $2$ . Construct a differential equation for $v$ in terms of time $t$ , including the value of the constant.
(4)
(Total for Question 2 is 4 marks)
3.
The radius $r$ metres of a circular oil patch increases at a rate inversely proportional to $r$ . When $r=6$ , the radius is increasing at $0.5$ metres per minute. Construct a differential equation for $r$ in terms of time $t$ minutes, including the constant of proportionality.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Demand $D$ is modelled as a function of price $p$ . The rate of decrease of demand with respect to price is proportional to $D$ and inversely proportional to $p^2$ . When $p=5$ and $D=800$ , $\frac{\mathrm dD}{\mathrm dp}=-64$ . Construct the differential equation, determining its constant.
(5)
(Total for Question 1 is 5 marks)
2.
Water enters a reservoir at a constant rate of $30\,\text{m}^3\text{h}^{-1}$ . Water leaves at a rate proportional to $\sqrt V$ , where $V$ is the volume in $\text{m}^3$ . When $V=100$ , the volume is decreasing at $10\,\text{m}^3\text{h}^{-1}$ . Construct a differential equation for $V$ in terms of time $t$ , determining the constant of proportionality.
(5)
(Total for Question 2 is 5 marks)
3.
A heater supplies energy so that the temperature $T$ would increase at a constant rate $a$ . Cooling occurs at a rate proportional to the amount by which $T$ exceeds the room temperature of $10^\circ\text{C}$ . When $T=20$ , $\dfrac{\mathrm dT}{\mathrm dt}=8$ ; when $T=50$ , $\dfrac{\mathrm dT}{\mathrm dt}=-4$ . Construct the differential equation, determining both constants, and find the equilibrium temperature.
(6)
(Total for Question 3 is 6 marks)
4.
A tank contains $500$ litres of well-stirred brine. Brine containing $0.02$ kg of salt per litre enters at $3$ litres per minute, and the mixture leaves at the same volume rate. Let $S$ be the mass of salt in the tank, in kg, after $t$ minutes. Construct a differential equation for $S$ , and find the equilibrium mass of salt.
(5)
(Total for Question 4 is 5 marks)
5.
A drug is infused at a rate of $6t$ mg per hour, where $t$ is the time in hours. The body removes the drug at a rate proportional to the current mass $M$ mg. At $t=2$ , $M=20$ and $\dfrac{\mathrm dM}{\mathrm dt}=7$ . Construct the differential equation, determining the positive proportionality constant. Find also the relation between $M$ and $t$ when the input and removal rates are instantaneously equal.
(5)
(Total for Question 5 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

7.1 · Derivative of f(x) as tangent gradient and as a limit; rate of change; sketch the gradient function; first-principles differentiation for small integer powers of x, sin x, cos x; second derivatives; convexity, concavity, inflection.

Tier 1 · Easy

Mark scheme for 7.1 Tier 1 · Easy
Question	Scheme	Marks
1	$f'(x)=2x-3$	4
(4 marks)		4
Notes
$f(x+h)-f(x)=(x+h)^2-3(x+h)-(x^2-3x)=2xh+h^2-3h$ . Hence $\frac{f(x+h)-f(x)}{h}=2x+h-3$ . Taking the limit as $h\to0$ gives $f'(x)=2x-3$ .
2	$f''(x)=12x^2-8$ The curve is locally convex at $x=1$	3
(3 marks)		3
Notes
$f'(x)=4x^3-8x$ , so $f''(x)=12x^2-8$ . Since $f''(1)=4>0$ , the curve is locally convex at $x=1$ .

Tier 2 · Standard

Mark scheme for 7.1 Tier 2 · Standard
Question	Scheme	Marks
1	$f'(a)=2a+3$	5
(5 marks)		5
Notes
By first principles, $f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ . Here $f(a+h)-f(a)=(a+h)^2+3(a+h)-(a^2+3a)=2ah+h^2+3h$ . For $h\neq0$ , division by $h$ gives $2a+h+3$ . Taking the limit as $h\to0$ gives $f'(a)=2a+3$ .
2	$f'(x)=3x^2+2$	5
(5 marks)		5
Notes
By first principles, $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}h$ . Expanding gives $f(x+h)-f(x)=3x^2h+3xh^2+h^3+2h$ . For $h\neq0$ , the quotient is $3x^2+3xh+h^2+2$ . Taking the limit as $h\to0$ gives $f'(x)=3x^2+2$ .
3	$\left.\dfrac{\mathrm dV}{\mathrm dx}\right\|_{x=2}=12\,\text{cm}^2$ Estimated increase $0.36\,\text{cm}^3$	4
(4 marks)		4
Notes
By first principles, $\left.\dfrac{\mathrm dV}{\mathrm dx}\right\|_{x=2}=\lim_{h\to0}\dfrac{(2+h)^3-2^3}{h}$ . Expanding the numerator gives $12h+6h^2+h^3$ , so for $h\neq0$ the quotient is $12+6h+h^2$ . Its limit as $h\to0$ is $12\,\text{cm}^2$ . For an increase of $0.03$ cm in edge length, $\Delta V\approx12(0.03)=0.36\,\text{cm}^3$ .

Tier 3 · Hard

Mark scheme for 7.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\frac{\mathrm d}{\mathrm dx}(\sin x)=\cos x$ $\frac{\mathrm d}{\mathrm dx}(\cos x)=-\sin x$	8
(8 marks)		8
Notes
For sine, $\frac{\sin(x+h)-\sin x}{h}=\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}$ , whose limit is $\cos x$ . For cosine, $\frac{\cos(x+h)-\cos x}{h}=\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}$ , whose limit is $-\sin x$ .
2	Increasing for $x>3$ Decreasing for $x<0$ and $0<x<3$ A stationary point of inflection at $x=0$ and a local minimum at $x=3$	6
(6 marks)		6
Notes
Since $x^2\geq0$ , the sign of $f'(x)=x^2(x-3)$ is negative for $x<3$ except at $x=0$ , and positive for $x>3$ . Hence $f$ decreases on either side of $0$ up to $3$ , then increases. Also $f''(x)=3x^2-6x=3x(x-2)$ , which changes sign at $x=0$ , so the stationary point there is an inflection. At $x=3$ , $f'$ changes from negative to positive (and $f''(3)=9>0$ ), giving a local minimum.
3	$f'(x)=3x^2+c$ $c=-5$ Tangent $y+2=7(x-2)$ , or $y=7x-16$	6
(6 marks)		6
Notes
From first principles, $f'(x)=\lim_{h\to0}\dfrac{(x+h)^3+c(x+h)-(x^3+cx)}{h}$ . For $h\neq0$ , expansion and division by $h$ give $3x^2+3xh+h^2+c$ , whose limit is $3x^2+c$ . The given line has gradient $-1/7$ , so the perpendicular tangent has gradient $7$ . Hence $f'(2)=12+c=7$ , giving $c=-5$ . The point on the curve is $(2,f(2))=(2,-2)$ , so the tangent is $y+2=7(x-2)$ .
4	Gradient at $x=a$ is $2a+p$ $p=-6$ Vertex $(3,-2)$ Tangent at the vertex: $y=-2$	7
(7 marks)		7
Notes
From first principles, the gradient at $x=a$ is $\lim_{h\to0}\dfrac{(a+h)^2+p(a+h)+7-(a^2+pa+7)}h$ . For $h\neq0$ , the quotient simplifies to $2a+h+p$ , whose limit is $2a+p$ . At $x=2$ , the point is $(2,11+2p)$ and the gradient is $4+p$ . Since the tangent also passes through $(4,-5)$ , $-5-(11+2p)=2(4+p)$ , giving $p=-6$ . Thus $y=x^2-6x+7=(x-3)^2-2$ , so the vertex is $(3,-2)$ . The limit result gives gradient $2(3)-6=0$ there, proving that its tangent is horizontal; its equation is $y=-2$ .
5	Exactly two distinct points of inflection when $\|k\|>1$ For $k=\dfrac54$ : points of inflection $\left(\dfrac12,\dfrac{15}{16}\right)$ and $(2,0)$ Convex for $x<\dfrac12$ and $x>2$ ; concave for $\dfrac12<x<2$	7
(7 marks)		7
Notes
$f''(x)=12x^2-24kx+12=12(x^2-2kx+1)$ . This quadratic has two distinct real roots exactly when its discriminant $4(k^2-1)$ is positive, so $\|k\|>1$ . Both roots are then simple, and the upward-opening quadratic changes sign at each, so both give points of inflection. When $k=5/4$ , $f''(x)=12(x-1/2)(x-2)$ , giving inflection values $x=1/2$ and $x=2$ . Substitution in $f(x)=x^4-5x^3+6x^2$ gives $f(1/2)=15/16$ and $f(2)=0$ . The sign of $f''$ is positive outside the two roots and negative between them, so the curve is convex for $x<1/2$ and $x>2$ , and concave for $1/2<x<2$ .

7.2 · Differentiate xⁿ for rational n, and related sums, differences and constant multiples; differentiate e^(kx), a^(kx), sin kx, cos kx, tan kx; understand and use the derivative of ln x.

Tier 1 · Easy

Mark scheme for 7.2 Tier 1 · Easy
Question	Scheme	Marks
1	$\frac{\mathrm dy}{\mathrm dx}=\frac{15}{2}x^{1/2}+x^{-3/2}$	3
(3 marks)		3
Notes
Apply the power rule term by term: $5\times\frac32x^{1/2}-2\times(-\frac12)x^{-3/2}=\frac{15}{2}x^{1/2}+x^{-3/2}$ .
2	$\dfrac{\mathrm dy}{\mathrm dx}=-15e^{-3x}+\dfrac2x$	2
(2 marks)		2
Notes
The exponential term differentiates to $5(-3)e^{-3x}$ and $\ln x$ differentiates to $1/x$ . Therefore $\dfrac{\mathrm dy}{\mathrm dx}=-15e^{-3x}+2/x$ .

Tier 2 · Standard

Mark scheme for 7.2 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{15}{2}x^{3/2}+2x^{-3/2}+2^x\ln2$	4
(4 marks)		4
Notes
Apply the power rule to the first two terms: $3(5/2)x^{3/2}=(15/2)x^{3/2}$ and $-4(-1/2)x^{-3/2}=2x^{-3/2}$ . Also $\dfrac{\mathrm d}{\mathrm dx}(2^x)=2^x\ln2$ . Adding the terms gives the stated derivative.
2	$f'(x)=3(\ln5)5^{3x}-8\cos(4x)-\dfrac32x^{-5/2}$	4
(4 marks)		4
Notes
Use $\dfrac{\mathrm d}{\mathrm dx}(a^{kx})=k(\ln a)a^{kx}$ , the chain rule for the sine term and the power rule. This gives $f'(x)=3(\ln5)5^{3x}-8\cos(4x)-(3/2)x^{-5/2}$ .
3	$y-\left(2+e^{-1}\right)=\left(4-e^{-1}\right)(x-1)$ (or $y=(4-e^{-1})x-2+2e^{-1}$ )	4
(4 marks)		4
Notes
Differentiating term by term gives $\dfrac{\mathrm dy}{\mathrm dx}=3x^{1/2}-e^{-x}+1/x$ . At $x=1$ , the gradient is $4-e^{-1}$ and the point on the curve is $\left(1,2+e^{-1}\right)$ . Therefore the tangent is $y-(2+e^{-1})=(4-e^{-1})(x-1)$ .

Tier 3 · Hard

Mark scheme for 7.2 Tier 3 · Hard
Question	Scheme	Marks
1	$f'(x)=2(\ln3)3^{2x}+5\sec^2(5x)+6\sin(3x)$ $f'(0)=2\ln3+5$	5
(5 marks)		5
Notes
Use the exponential and trigonometric derivatives: $f'(x)=2(\ln3)3^{2x}+5\sec^2(5x)+6\sin(3x)$ . At $x=0$ , $3^0=1$ , $\sec^20=1$ and $\sin0=0$ , giving $f'(0)=2\ln3+5$ .
2	$f'(x)=3(\ln2)2^{3x}+8\cos(2x)-\dfrac1x$ $f'(\pi/4)=3(\ln2)2^{3\pi/4}-\dfrac4\pi$	5
(5 marks)		5
Notes
Differentiate term by term to obtain $f'(x)=3(\ln2)2^{3x}+8\cos(2x)-1/x$ . At $x=\pi/4$ , $\cos(\pi/2)=0$ and $1/x=4/\pi$ , giving the stated exact value.
3	$A=2$ and $B=3$ $f'(9)=\dfrac{28}{3}$	6
(6 marks)		6
Notes
Differentiating gives $f'(x)=\dfrac{3A}{2}x^{1/2}+\dfrac Bx+\dfrac\pi2\cos(\pi x/2)$ . At $x=1$ , the cosine term is zero, giving $3A/2+B=6$ . At $x=4$ , $\cos(2\pi)=1$ , so comparison with the given value gives $3A+B/4=27/4$ . Solving these simultaneous equations gives $A=2$ and $B=3$ . At $x=9$ , $\cos(9\pi/2)=0$ , so $f'(9)=3\sqrt9+3/9=9+1/3=28/3$ .
4	$a=-\dfrac{\ln2}{\pi}$ Tangent $y-2=3\ln2(x-1)$	6
(6 marks)		6
Notes
$f'(x)=(\ln2)2^x+a\pi\cos(\pi x)$ . The stationary condition at $x=0$ gives $\ln2+a\pi=0$ , so $a=-\ln2/\pi$ . At $x=1$ , $f(1)=2$ and $f'(1)=2\ln2+a\pi\cos\pi=2\ln2+\ln2=3\ln2$ . The tangent therefore has the stated equation.
5	$\left(\dfrac\pi3,\dfrac{3\sqrt3}{2}\right)$ , $(\pi,0)$ and $\left(\dfrac{5\pi}{3},-\dfrac{3\sqrt3}{2}\right)$	6
(6 marks)		6
Notes
$\mathrm dy/\mathrm dx=2\cos(2x)+2\cos x$ . A stationary point satisfies $\cos(2x)+\cos x=0$ . With $c=\cos x$ and $\cos(2x)=2c^2-1$ , this becomes $2c^2+c-1=(2c-1)(c+1)=0$ . Thus $\cos x=1/2$ or $\cos x=-1$ . The complete solutions in the interval are $x=\pi/3,\pi,5\pi/3$ . Substitution in the original equation gives the three stated coordinates.

7.3 · Apply differentiation to find gradients, tangents and normals, maxima and minima and stationary points, points of inflection; identify where functions are increasing or decreasing.

Tier 1 · Easy

Mark scheme for 7.3 Tier 1 · Easy
Question	Scheme	Marks
1	Tangent $y=5x-1$ Normal $y-4=-\frac15(x-1)$	4
(4 marks)		4
Notes
At $x=1$ , $y=1+3=4$ . Since $\frac{\mathrm dy}{\mathrm dx}=2x+3$ , the tangent gradient is $5$ , giving $y-4=5(x-1)$ and hence $y=5x-1$ . The normal gradient is $-\frac15$ , so $y-4=-\frac15(x-1)$ .
2	$y=-16$	3
(3 marks)		3
Notes
At $x=2$ , $y=8-24=-16$ . Also $\dfrac{\mathrm dy}{\mathrm dx}=3x^2-12$ , which is $0$ at $x=2$ . The tangent is therefore the horizontal line $y=-16$ .

Tier 2 · Standard

Mark scheme for 7.3 Tier 2 · Standard
Question	Scheme	Marks
1	Tangent: $y-5=\dfrac72(x-2)$ , or any algebraically equivalent equation such as $y=\dfrac72x-2$ Normal: $y-5=-\dfrac27(x-2)$ , or any algebraically equivalent equation such as $y=-\dfrac27x+\dfrac{39}{7}$	5
(5 marks)		5
Notes
At $x=2$ , $y=2^2+2/2=5$ . Differentiate: $dy/dx=2x-2/x^2$ , so the tangent gradient is $4-1/2=7/2$ . Thus the tangent is $y-5=\tfrac72(x-2)$ . The normal gradient is the negative reciprocal, $-2/7$ , so the normal is $y-5=-\tfrac27(x-2)$ .
2	Local maximum $(-1,10)$ Local minimum $(3,-22)$	5
(5 marks)		5
Notes
$f'(x)=3x^2-6x-9=3(x+1)(x-3)$ , so stationary points occur at $x=-1$ and $x=3$ . Their coordinates are $(-1,10)$ and $(3,-22)$ . Since $f''(x)=6x-6$ , $f''(-1)=-12<0$ gives a local maximum and $f''(3)=12>0$ gives a local minimum.
3	$(-2,-2)$ and $(2,2)$	4
(4 marks)		4
Notes
The given line has gradient $9$ . For the curve, $\dfrac{\mathrm dy}{\mathrm dx}=3x^2-3$ . Parallel tangents therefore satisfy $3x^2-3=9$ , so $x^2=4$ and $x=\pm2$ . Substitution into $y=x^3-3x$ gives $y=-2$ when $x=-2$ and $y=2$ when $x=2$ .

Tier 3 · Hard

Mark scheme for 7.3 Tier 3 · Hard
Question	Scheme	Marks
1	Maximum volume $256\,\text{cm}^3$ , attained when $x=4$	7
(7 marks)		7
Notes
Differentiate the product: $V'=(12-x)^2-2x(12-x)=(12-x)(12-3x)$ . The interior stationary point is $x=4$ ; $x=12$ is outside the open domain as a stationary endpoint factor. Since $V'$ changes from positive to negative at $x=4$ , this is a maximum. Thus $V(4)=4(8)^2=256\,\text{cm}^3$ . Also $V$ tends to $0$ at both ends of the domain, confirming the global maximum.
2	$P=(2\sqrt3,2\sqrt3)$ Minimum distance $2\sqrt6$	6
(6 marks)		6
Notes
The squared distance is $D^2=x^2+y^2=x^2+144x^{-2}$ . Differentiating gives $\dfrac{\mathrm d(D^2)}{\mathrm dx}=2x-288x^{-3}$ . Setting this to zero gives $x^4=144$ , so $x=2\sqrt3$ because $x>0$ . Then $y=12/x=2\sqrt3$ . Also the second derivative is $2+864/x^4>0$ , so this is a minimum. Hence $D=\sqrt{12+12}=2\sqrt6$ .
3	Dimensions $4$ by $8$ Maximum area $32$ square units	6
(6 marks)		6
Notes
The width is $2x$ and the height is $12-x^2$ , so $A=2x(12-x^2)=24x-2x^3$ . Then $\dfrac{\mathrm dA}{\mathrm dx}=24-6x^2$ , which is zero at $x=2$ in the stated domain. Also $\dfrac{\mathrm d^2A}{\mathrm dx^2}=-12x<0$ there, so this gives a maximum. The width is $2x=4$ , the height is $12-x^2=8$ , and the maximum area is $4\times8=32$ square units.
4	$k<0$ : no stationary points $k=0$ : one stationary point of inflection at $(0,2)$ $k>0$ : a local maximum at $x=-\sqrt{k}$ and a local minimum at $x=\sqrt{k}$ When $k=4$ , local maximum $(-2,18)$ and local minimum $(2,-14)$	7
(7 marks)		7
Notes
$\mathrm dy/\mathrm dx=3(x^2-k)$ . If $k<0$ , this derivative is always positive. If $k=0$ , it is zero only at $x=0$ and does not change sign, giving a stationary point of inflection at $(0,2)$ . If $k>0$ , the roots are $x=\pm\sqrt{k}$ ; the derivative changes from positive to negative at the negative root and from negative to positive at the positive root. For $k=4$ , substitution of $x=-2$ and $x=2$ gives $y=18$ and $y=-14$ respectively.
5	Points of contact $(-2,4)$ and $(2,4)$ Tangents $y=-4x-4$ and $y=4x-4$ Acute angle $=\tan^{-1}\!\left(\dfrac8{15}\right)$	6
(6 marks)		6
Notes
At the point $(a,a^2)$ , the parabola has gradient $2a$ , so its tangent is $y-a^2=2a(x-a)$ , or $y=2ax-a^2$ . Passing through $(0,-4)$ requires $-4=-a^2$ , hence $a=\pm2$ . These give the contact points $(-2,4)$ and $(2,4)$ and the two stated tangent equations. If $\theta$ is the acute angle between lines of gradients $4$ and $-4$ , then $\tan\theta=\left\|\dfrac{4-(-4)}{1+(4)(-4)}\right\|=8/15$ . Therefore $\theta=\tan^{-1}(8/15)$ .

7.4 · Differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions.

Tier 1 · Easy

Mark scheme for 7.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\frac{\mathrm dy}{\mathrm dx}=e^{3x}(2x+3x^2)$	3
(3 marks)		3
Notes
Apply the product rule: $\frac{\mathrm dy}{\mathrm dx}=(2x)e^{3x}+x^2(3e^{3x})=e^{3x}(2x+3x^2)$ .
2	$\dfrac{\mathrm dy}{\mathrm dx}=24x(3x^2-1)^3$	2
(2 marks)		2
Notes
Apply the chain rule: multiply $4(3x^2-1)^3$ by the inner derivative $6x$ . This gives $24x(3x^2-1)^3$ .

Tier 2 · Standard

Mark scheme for 7.4 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{\mathrm dr}{\mathrm dt}=\dfrac13\,\text{cm s}^{-1}$	4
(4 marks)		4
Notes
For $V=\tfrac43\pi r^3$ , the chain rule gives $dV/dt=4\pi r^2\,dr/dt$ . Substitute $dV/dt=12\pi$ and $r=3$ : $12\pi=4\pi(3^2)\,dr/dt=36\pi\,dr/dt$ . Therefore $dr/dt=1/3\,\text{cm s}^{-1}$ .
2	$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2(x^2+4)-2x(2x+1)\ln(2x+1)}{(2x+1)(x^2+4)^2}$	4
(4 marks)		4
Notes
Let $u=\ln(2x+1)$ and $v=x^2+4$ . Then $u'=2/(2x+1)$ and $v'=2x$ . The quotient rule gives $\dfrac{u'v-uv'}{v^2}$ . Multiplying the numerator and denominator by $2x+1$ gives $\dfrac{2(x^2+4)-2x(2x+1)\ln(2x+1)}{(2x+1)(x^2+4)^2}$ .
3	$\dfrac{\mathrm dy}{\mathrm dt}=-\dfrac34$ units per second $P$ is moving downwards	4
(4 marks)		4
Notes
Differentiate $x^2+4y^2=100$ with respect to time: $2x\dfrac{\mathrm dx}{\mathrm dt}+8y\dfrac{\mathrm dy}{\mathrm dt}=0$ . At $P=(6,4)$ with $\dfrac{\mathrm dx}{\mathrm dt}=2$ , this gives $2(6)(2)+8(4)\dfrac{\mathrm dy}{\mathrm dt}=0$ . Hence $\dfrac{\mathrm dy}{\mathrm dt}=-24/32=-3/4$ units per second. The negative sign shows that $P$ is moving downwards.

Tier 3 · Hard

Mark scheme for 7.4 Tier 3 · Hard
Question	Scheme	Marks
1	(a) $\frac{\mathrm dr}{\mathrm dt}=\frac23\,\text{cm s}^{-1}$ (b) $g'(3)=\frac14$	7
(7 marks)		7
Notes
(a) Differentiate with respect to time: $\frac{\mathrm dV}{\mathrm dt}=4\pi r^2\frac{\mathrm dr}{\mathrm dt}$ . At $r=3$ , $24\pi=36\pi\frac{\mathrm dr}{\mathrm dt}$ , so $\frac{\mathrm dr}{\mathrm dt}=\frac23\,\text{cm s}^{-1}$ . (b) Since $g=f^{-1}$ , $g'(3)=1/f'(g(3))$ . Now $f'(x)=3x^2+1$ and $g(3)=1$ , so $g'(3)=1/f'(1)=\frac14$ .
2	$\dfrac{\mathrm dr}{\mathrm dt}=\dfrac94\,\text{cm s}^{-1}$ $\dfrac{\mathrm dh}{\mathrm dt}=\dfrac92\,\text{cm s}^{-1}$	6
(6 marks)		6
Notes
Using $h=2r$ , $V=\frac23\pi r^3$ , so $\dfrac{\mathrm dV}{\mathrm dt}=2\pi r^2\dfrac{\mathrm dr}{\mathrm dt}$ . When $h=4$ , $r=2$ . Hence $18\pi=2\pi(2^2)\dfrac{\mathrm dr}{\mathrm dt}$ , giving $\dfrac{\mathrm dr}{\mathrm dt}=9/4$ . Since $h=2r$ , differentiating with respect to time gives $\dfrac{\mathrm dh}{\mathrm dt}=2\dfrac{\mathrm dr}{\mathrm dt}=9/2$ .
3	$\dfrac{\mathrm dw}{\mathrm dt}=5\,\text{cm s}^{-1}$ $\dfrac{\mathrm dl}{\mathrm dt}=10\,\text{cm s}^{-1}$ $\dfrac{\mathrm dP}{\mathrm dt}=30\,\text{cm s}^{-1}$	6
(6 marks)		6
Notes
Using $l=2w+3$ , the area is $A=wl=2w^2+3w$ . Differentiating with respect to time gives $\dfrac{\mathrm dA}{\mathrm dt}=(4w+3)\dfrac{\mathrm dw}{\mathrm dt}$ . At $w=4$ , $95=19\,\dfrac{\mathrm dw}{\mathrm dt}$ , so $\dfrac{\mathrm dw}{\mathrm dt}=5$ . Differentiating $l=2w+3$ gives $\dfrac{\mathrm dl}{\mathrm dt}=10$ . Since $P=2l+2w$ , $\dfrac{\mathrm dP}{\mathrm dt}=2(10)+2(5)=30\,\text{cm s}^{-1}$ .
4	$\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{3}{40}$ rad s $^{-1}$ The top moves downwards at $\dfrac9{20}$ m s $^{-1}$	6
(6 marks)		6
Notes
If the foot is $x$ metres from the wall, $x=10\cos\theta$ . At $x=6$ , the height is $8$ and hence $\sin\theta=4/5$ , $\cos\theta=3/5$ . Differentiating gives $\mathrm dx/\mathrm dt=-10\sin\theta\,\mathrm d\theta/\mathrm dt$ . Thus $0.6=-8\,\mathrm d\theta/\mathrm dt$ , so $\mathrm d\theta/\mathrm dt=-3/40$ . With height $y=10\sin\theta$ , $\mathrm dy/\mathrm dt=10\cos\theta\,\mathrm d\theta/\mathrm dt=6(-3/40)=-9/20$ , so the top moves downwards.
5	$\dfrac{\mathrm dy}{\mathrm dx}=e^{-x}(x^2+1)^2(-x^2+6x-1)$ $x=3-2\sqrt2$ is a local minimum $x=3+2\sqrt2$ is a local maximum	6
(6 marks)		6
Notes
Using the product and chain rules, $y'=6x(x^2+1)^2e^{-x}-(x^2+1)^3e^{-x}=e^{-x}(x^2+1)^2(-x^2+6x-1)$ . The positive factors never vanish, so stationary points satisfy $x^2-6x+1=0$ , giving $x=3\pm2\sqrt2$ . The quadratic factor in $y'$ is negative before the smaller root, positive between the roots and negative after the larger root. Therefore the smaller root is a local minimum and the larger root a local maximum.

7.5 · Differentiate simple functions and relations defined implicitly or parametrically, for first derivative only.

Tier 1 · Easy

Mark scheme for 7.5 Tier 1 · Easy
Question	Scheme	Marks
1	$\frac{\mathrm dy}{\mathrm dx}=-\frac{x}{y}$ for $y\ne0$	3
(3 marks)		3
Notes
Differentiate both sides with respect to $x$ : $2x+2y\frac{\mathrm dy}{\mathrm dx}=0$ . For $y\ne0$ , rearranging gives $\frac{\mathrm dy}{\mathrm dx}=-\frac{x}{y}$ . At $y=0$ the curve instead has a vertical tangent.
2	$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{5}{2t}$ for $t\neq0$	3
(3 marks)		3
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=2t$ and $\dfrac{\mathrm dy}{\mathrm dt}=5$ . Therefore $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}=5/(2t)$ , provided $t\neq0$ .

Tier 2 · Standard

Mark scheme for 7.5 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2x+y}{x+2y}$ Tangent $y-2=-\dfrac45(x-1)$ , or any algebraically equivalent equation such as $4x+5y-14=0$	5
(5 marks)		5
Notes
Differentiate implicitly: $2x+(x\,dy/dx+y)+2y\,dy/dx=0$ . Collecting derivative terms gives $(x+2y)dy/dx=-(2x+y)$ , so $dy/dx=-(2x+y)/(x+2y)$ . At $(1,2)$ the gradient is $-(2+2)/(1+4)=-4/5$ , hence the tangent is $y-2=-\tfrac45(x-1)$ .
2	$t=2$ Tangent $y=-4$ Normal $x=9$	5
(5 marks)		5
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=3t^2$ and $\dfrac{\mathrm dy}{\mathrm dt}=2t-4$ . A horizontal tangent requires $2t-4=0$ , so $t=2$ ; here $\mathrm dx/\mathrm dt=12\neq0$ . The point is $(2^3+1,2^2-4(2))=(9,-4)$ . Hence the horizontal tangent is $y=-4$ and the normal is the vertical line $x=9$ .
3	$y-\dfrac{3\sqrt2}{2}=-\dfrac32(x-\sqrt2)$	4
(4 marks)		4
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=-2\sin t$ and $\dfrac{\mathrm dy}{\mathrm dt}=3\cos t$ , so $\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3\cos t}{2\sin t}$ . At $t=\pi/4$ , the gradient is $-3/2$ and the point is $(\sqrt2,3\sqrt2/2)$ . Hence the tangent is $y-3\sqrt2/2=-\tfrac32(x-\sqrt2)$ .

Tier 3 · Hard

Mark scheme for 7.5 Tier 3 · Hard
Question	Scheme	Marks
1	Tangent $y-\frac32=\frac53\left(x-\frac52\right)$ Normal $y-\frac32=-\frac35\left(x-\frac52\right)$	7
(7 marks)		7
Notes
$\frac{\mathrm dx}{\mathrm dt}=1-\frac1{t^2}$ and $\frac{\mathrm dy}{\mathrm dt}=1+\frac1{t^2}$ . At $t=2$ , these are $\frac34$ and $\frac54$ , so the tangent gradient is $\frac53$ and the normal gradient is $-\frac35$ . The point is $(\frac52,\frac32)$ , giving the two stated equations.
2	Area of triangle $OAB=\dfrac{49}{60}$ square units	7
(7 marks)		7
Notes
Implicit differentiation gives $2x+y+(x+4y)\dfrac{\mathrm dy}{\mathrm dx}=0$ . At $(2,1)$ , the tangent gradient is $-5/6$ , so the normal gradient is $6/5$ . Its equation is $y-1=\frac65(x-2)$ . Setting $y=0$ gives the $x$ -intercept $A=(7/6,0)$ ; setting $x=0$ gives $B=(0,-7/5)$ . Therefore the area is $\frac12\left\|\frac76\cdot-\frac75\right\|=49/60$ square units.
3	$t=1$ , with tangent $y=x-1$ $t=-\dfrac13$ , with tangent $y=x+\dfrac5{27}$	6
(6 marks)		6
Notes
$\dfrac{\mathrm dx}{\mathrm dt}=2t+1$ and $\dfrac{\mathrm dy}{\mathrm dt}=3t^2$ , so $\dfrac{\mathrm dy}{\mathrm dx}=3t^2/(2t+1)$ where $t\neq-1/2$ . Setting this equal to $1$ gives $3t^2-2t-1=0=(3t+1)(t-1)$ , so $t=1$ or $t=-1/3$ . At $t=1$ the point is $(2,1)$ , giving $y=x-1$ . At $t=-1/3$ the point is $(-2/9,-1/27)$ , giving $y+1/27=x+2/9$ , or $y=x+5/27$ .
4	$t=-1$ gives tangent $y=-x$ $t=1$ gives tangent $y=x$ The tangents are perpendicular	6
(6 marks)		6
Notes
At the origin, $t^2-1=0$ , so $t=\pm1$ ; both also give $t^3-t=0$ . Now $\mathrm dx/\mathrm dt=2t$ and $\mathrm dy/\mathrm dt=3t^2-1$ , so $\mathrm dy/\mathrm dx=(3t^2-1)/(2t)$ . The gradients are $-1$ at $t=-1$ and $1$ at $t=1$ . Since both lines pass through the origin, their equations are $y=-x$ and $y=x$ ; their gradient product is $-1$ , so they are perpendicular.
5	Horizontal tangents at $(-\sqrt6,\sqrt6)$ and $(\sqrt6,-\sqrt6)$ Vertical tangents at $(-3\sqrt2,\sqrt2)$ and $(3\sqrt2,-\sqrt2)$	7
(7 marks)		7
Notes
Implicit differentiation gives $2x+2y+(2x+6y)\,\mathrm dy/\mathrm dx=0$ , so $\mathrm dy/\mathrm dx=-(x+y)/(x+3y)$ . Horizontal tangents require $x+y=0$ with $x+3y\neq0$ . Substituting $x=-y$ into the curve gives $2y^2=12$ , producing the two stated points. Vertical tangents require $x+3y=0$ with $x+y\neq0$ . Substituting $x=-3y$ gives $6y^2=12$ , producing the other two points.

7.6 · Construct simple differential equations in pure mathematics and in context (contexts may include kinematics, population growth and modelling the relationship between price and demand).

Tier 1 · Easy

Mark scheme for 7.6 Tier 1 · Easy
Question	Scheme	Marks
1	$\frac{\mathrm dv}{\mathrm dt}=-kv$ , where $k>0$	2
(2 marks)		2
Notes
Acceleration is $\frac{\mathrm dv}{\mathrm dt}$ . Proportionality to $v$ gives magnitude $kv$ with $k>0$ , and opposition to the motion supplies the minus sign: $\frac{\mathrm dv}{\mathrm dt}=-kv$ .
2	$\dfrac{\mathrm dy}{\mathrm dx}=kx^2$ , where $k$ is a constant	2
(2 marks)		2
Notes
The derivative $\dfrac{\mathrm dy}{\mathrm dx}$ represents the stated rate of change. Proportionality to $x^2$ therefore gives $\dfrac{\mathrm dy}{\mathrm dx}=kx^2$ for a constant $k$ .

Tier 2 · Standard

Mark scheme for 7.6 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{\mathrm dP}{\mathrm dt}=\dfrac{1}{5000}P(1000-P)$	4
(4 marks)		4
Notes
The statement gives $dP/dt=kP(1000-P)$ with $k>0$ . Using $P=200$ and $dP/dt=32$ gives $32=k(200)(800)=160000k$ , so $k=1/5000$ . Hence $dP/dt=P(1000-P)/5000$ .
2	$\dfrac{\mathrm dv}{\mathrm dt}=6-v^2$	4
(4 marks)		4
Notes
Write $\dfrac{\mathrm dv}{\mathrm dt}=6-kv^2$ . Using $v=2$ and acceleration $2$ gives $2=6-4k$ , so $k=1$ . Hence $\dfrac{\mathrm dv}{\mathrm dt}=6-v^2$ .
3	$\dfrac{\mathrm dr}{\mathrm dt}=\dfrac3r$	4
(4 marks)		4
Notes
Inverse proportionality gives $\dfrac{\mathrm dr}{\mathrm dt}=k/r$ for a positive constant $k$ . Using $r=6$ and $\dfrac{\mathrm dr}{\mathrm dt}=0.5$ gives $0.5=k/6$ , so $k=3$ . Therefore $\dfrac{\mathrm dr}{\mathrm dt}=3/r$ .

Tier 3 · Hard

Mark scheme for 7.6 Tier 3 · Hard
Question	Scheme	Marks
1	$\frac{\mathrm dD}{\mathrm dp}=-\frac{2D}{p^2}$	5
(5 marks)		5
Notes
The description gives $\frac{\mathrm dD}{\mathrm dp}=-\frac{kD}{p^2}$ with $k>0$ . Substitute the data: $-64=-\frac{k(800)}{25}=-32k$ , so $k=2$ . Hence $\frac{\mathrm dD}{\mathrm dp}=-\frac{2D}{p^2}$ .
2	$\dfrac{\mathrm dV}{\mathrm dt}=30-4\sqrt V$	5
(5 marks)		5
Notes
The net rate is inflow minus outflow, so $\dfrac{\mathrm dV}{\mathrm dt}=30-k\sqrt V$ . At $V=100$ , the net rate is $-10$ , giving $-10=30-10k$ and hence $k=4$ . Therefore $\dfrac{\mathrm dV}{\mathrm dt}=30-4\sqrt V$ .
3	$a=12$ and the cooling constant is $\dfrac25$ $\dfrac{\mathrm dT}{\mathrm dt}=12-\dfrac25(T-10)$ Equilibrium temperature $40^\circ\text{C}$	6
(6 marks)		6
Notes
Write $\dfrac{\mathrm dT}{\mathrm dt}=a-k(T-10)$ with $k>0$ . The two observations give $8=a-10k$ and $-4=a-40k$ . Subtracting gives $12=30k$ , so $k=2/5$ , and then $a=8+10(2/5)=12$ . Thus $\dfrac{\mathrm dT}{\mathrm dt}=12-(2/5)(T-10)$ . At equilibrium the derivative is zero, so $12=(2/5)(T-10)$ and $T=40^\circ\text{C}$ .
4	$\dfrac{\mathrm dS}{\mathrm dt}=0.06-\dfrac{3S}{500}$ Equilibrium mass $S=10$ kg	5
(5 marks)		5
Notes
Salt enters at $3(0.02)=0.06$ kg per minute. Because the mixture is well stirred and the volume stays at $500$ litres, its salt concentration is $S/500$ kg per litre, so salt leaves at $3S/500$ kg per minute. Therefore the net rate is $\mathrm dS/\mathrm dt=0.06-3S/500$ . At equilibrium this rate is zero, giving $S=10$ kg.
5	$\dfrac{\mathrm dM}{\mathrm dt}=6t-\dfrac14M$ The rates are instantaneously equal when $M=24t$	5
(5 marks)		5
Notes
Write $\mathrm dM/\mathrm dt=6t-kM$ with $k>0$ . The data give $7=6(2)-20k$ , so $20k=5$ and $k=1/4$ . Hence $\mathrm dM/\mathrm dt=6t-M/4$ . Equality of the input and removal rates requires $6t=M/4$ , which is equivalent to $M=24t$ .