7 Differentiation — 9MA0 spec coverage

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

4 marks

ORIGINAL

For $f(x)=x^2-3x$ , use the limit definition of the derivative to find $f'(x)$ .

Tier 2 · Standard

6 marks

ORIGINAL

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.

Tier 3 · Hard

8 marks

ORIGINAL

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 $\lim_{h\to0}\frac{\cos h-1}{h}=0$ .

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

3 marks

ORIGINAL

Differentiate $y=5x^{3/2}-2x^{-1/2}$ with respect to $x$ .

Tier 2 · Standard

4 marks

ORIGINAL

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

Tier 3 · Hard

5 marks

ORIGINAL

Given $f(x)=3^{2x}+\tan(5x)-2\cos(3x)$ , find $f'(x)$ and hence find the exact value of $f'(0)$ .

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

4 marks

ORIGINAL

The curve $y=x^2+3x$ is considered at the point where $x=1$ . Find the equations of the tangent and the normal.

Tier 2 · Standard

7 marks

ORIGINAL

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

Tier 3 · Hard

7 marks

ORIGINAL

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.

7.4

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

3 marks

ORIGINAL

Differentiate $y=x^2e^{3x}$ .

Tier 2 · Standard

9 marks

ORIGINAL

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

Tier 3 · Hard

7 marks

ORIGINAL

(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 $g'(3)$ .

7.5

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

3 marks

ORIGINAL

The curve obeys $x^2+y^2=25$ . By implicit differentiation, obtain its gradient at a general point $(x,y)$ .

Tier 2 · Standard

5 marks

ORIGINAL

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

Tier 3 · Hard

7 marks

ORIGINAL

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 $t=2$ .

7.6

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

2 marks

ORIGINAL

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 $t$ .

Tier 2 · Standard

4 marks

ORIGINAL

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.

Tier 3 · Hard

5 marks

ORIGINAL

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.