7 Differentiation — printable coverage pack

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.

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.

Tier 1 · Easy

1. For $f(x)=x^2-3x$ , use the limit definition of the derivative to find $f'(x)$ .[4 marks]
Answer
- $f'(x)=2x-3$
Method: $f(x+h)-f(x)=(x+h)^2-3(x+h)-(x^2-3x)=2xh+h^2-3h$ . Hence $\frac{f(x+h)-f(x)}{h}=2x+h-3$ . Taking the limit as $h\to0$ gives $f'(x)=2x-3$ .

Tier 2 · Standard

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

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 marks]
Answer
- $\frac{\mathrm d}{\mathrm dx}(\sin x)=\cos x$
- $\frac{\mathrm d}{\mathrm dx}(\cos x)=-\sin x$
Method: 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 $-\sin x$ .

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.

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 $a^{kx}$ .

Tier 1 · Easy

1. Differentiate $y=5x^{3/2}-2x^{-1/2}$ with respect to $x$ .[3 marks]
Answer
- $\frac{\mathrm dy}{\mathrm dx}=\frac{15}{2}x^{1/2}+x^{-3/2}$
Method: Apply the power rule term by term: $5\times\frac32x^{1/2}-2\times(-\frac12)x^{-3/2}=\frac{15}{2}x^{1/2}+x^{-3/2}$ .

Tier 2 · Standard

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

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 marks]
Answer
- $f'(x)=2(\ln3)3^{2x}+5\sec^2(5x)+6\sin(3x)$
- $f'(0)=2\ln3+5$
Method: 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 $f'(0)=2\ln3+5$ .

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.

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.

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 marks]
Answer
- Tangent $y=5x-1$
- Normal $y-4=-\frac15(x-1)$
Method: 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 $y-4=-\frac15(x-1)$ .

Tier 2 · Standard

1. For $f(x)=x^3-6x^2+9x+2$ , find and classify every stationary point. State the intervals on which $f$ is increasing.[7 marks]
Answer
- Local maximum $(1,6)$ and local minimum $(3,2)$
- Increasing for $x<1$ and $x>3$
Method: $f'(x)=3x^2-12x+9=3(x-1)(x-3)$ , so the stationary values are $x=1,3$ . The coordinates are $f(1)=6$ and $f(3)=2$ . Also $f''(x)=6x-12$ , so $f''(1)=-6$ gives a local maximum and $f''(3)=6$ gives a local minimum. The factorised derivative is positive outside the roots, so $f$ is increasing for $x<1$ and $x>3$ .

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 marks]
Answer
- Maximum volume $256\,\text{cm}^3$ , attained when $x=4$
Method: 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.

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

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.

Tier 1 · Easy

1. Differentiate $y=x^2e^{3x}$ .[3 marks]
Answer
- $\frac{\mathrm dy}{\mathrm dx}=e^{3x}(2x+3x^2)$
Method: Apply the product rule: $\frac{\mathrm dy}{\mathrm dx}=(2x)e^{3x}+x^2(3e^{3x})=e^{3x}(2x+3x^2)$ .

Tier 2 · Standard

1. (a) Differentiate $y=\frac{\sin(2x)}{(x^2+1)^3}$ , giving one fraction. (b) Differentiate $z=\sec(3x)-2\cot x+\cosec(2x)$ .[9 marks]
Answer
- (a) $\frac{\mathrm dy}{\mathrm dx}=\frac{2(x^2+1)\cos(2x)-6x\sin(2x)}{(x^2+1)^4}$
- (b) $\frac{\mathrm dz}{\mathrm dx}=3\sec(3x)\tan(3x)+2\cosec^2x-2\cosec(2x)\cot(2x)$
Method: (a) Take $u=\sin(2x)$ and $v=(x^2+1)^3$ . Then $u'=2\cos(2x)$ and $v'=6x(x^2+1)^2$ . 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.

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 marks]
Answer
- (a) $\frac{\mathrm dr}{\mathrm dt}=\frac23\,\text{cm s}^{-1}$
- (b) $g'(3)=\frac14$
Method: (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 $g'(3)=1/f'(1)=\frac14$ .

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

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 $y+x\frac{\mathrm dy}{\mathrm dx}$ .

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 marks]
Answer
- $\frac{\mathrm dy}{\mathrm dx}=-\frac{x}{y}$ for $y\ne0$
Method: 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.

Tier 2 · Standard

1. A curve has parametric equations $x=t^2+1$ and $y=t^3-3t$ . Find the equation of its tangent when $t=2$ .[5 marks]
Answer
- $y-2=\frac94(x-5)$
Method: $\frac{\mathrm dx}{\mathrm dt}=2t$ and $\frac{\mathrm dy}{\mathrm dt}=3t^2-3$ , so $\frac{\mathrm dy}{\mathrm dx}=\frac{3t^2-3}{2t}$ . At $t=2$ , the gradient is $\frac94$ . The point is $(5,2)$ , so the tangent is $y-2=\frac94(x-5)$ .

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 marks]
Answer
- Tangent $y-\frac32=\frac53\left(x-\frac52\right)$
- Normal $y-\frac32=-\frac35\left(x-\frac52\right)$
Method: $\frac{\mathrm dx}{\mathrm dt}=1-\frac1{t^2}$ 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.

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).

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.

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 marks]
Answer
- $\frac{\mathrm dv}{\mathrm dt}=-kv$ , where $k>0$
Method: 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: $\frac{\mathrm dv}{\mathrm dt}=-kv$ .

Tier 2 · Standard

1. 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.[4 marks]
Answer
- $\frac{\mathrm dP}{\mathrm dt}=\frac18(1200-P)$
Method: Write $\frac{\mathrm dP}{\mathrm dt}=k(1200-P)$ . Using $P=800$ and rate $50$ gives $50=k(400)$ , so $k=\frac18$ . Therefore $\frac{\mathrm dP}{\mathrm dt}=\frac18(1200-P)$ .

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 marks]
Answer
- $\frac{\mathrm dD}{\mathrm dp}=-\frac{2D}{p^2}$
Method: 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 $\frac{\mathrm dD}{\mathrm dp}=-\frac{2D}{p^2}$ .