Note AcademyNote Academy
Start

Mathematics (AS & A Level) 9709 · Chapter 12

Further differentiation

Papers 2 and 3. Derivatives of exponentials, logarithms and trigonometric functions, the product and quotient rules, and curves defined parametrically or implicitly. Paper 3 adds the inverse tangent.

Written for the 2026–2027 syllabus · 4 min read

Differentiation

Standard derivatives

yydydx\dfrac{\mathrm{d}y}{\mathrm{d}x}
ex\mathrm{e}^xex\mathrm{e}^x
ln⁡x\ln x1x\dfrac{1}{x}
sin⁡x\sin xcos⁡x\cos x
cos⁡x\cos x−sin⁡x-\sin x
tan⁡x\tan xsec⁡2x\sec^2 x
tan⁡−1x\tan^{-1} x (Paper 3)11+x2\dfrac{1}{1 + x^2}

Composites

The chain rule gives these general results:

yydydx\dfrac{\mathrm{d}y}{\mathrm{d}x}
ef(x)\mathrm{e}^{\mathrm{f}(x)}f′(x)ef(x)\mathrm{f}'(x)\mathrm{e}^{\mathrm{f}(x)}
ln⁡f(x)\ln \mathrm{f}(x)f′(x)f(x)\dfrac{\mathrm{f}'(x)}{\mathrm{f}(x)}
sin⁡f(x)\sin \mathrm{f}(x)f′(x)cos⁡f(x)\mathrm{f}'(x)\cos \mathrm{f}(x)
cos⁡f(x)\cos \mathrm{f}(x)−f′(x)sin⁡f(x)-\mathrm{f}'(x)\sin \mathrm{f}(x)
tan⁡f(x)\tan \mathrm{f}(x)f′(x)sec⁡2f(x)\mathrm{f}'(x)\sec^2 \mathrm{f}(x)

6x e3x26x3x2+12sec⁡22x2sin⁡xcos⁡x=sin⁡2x6x\,\mathrm{e}^{3x^2} \qquad \frac{6x}{3x^2 + 1} \qquad 2\sec^2 2x \qquad 2\sin x\cos x = \sin 2x

sin⁡2x\sin^2 x means (sin⁡x)2(\sin x)^2, so use the power rule first and multiply by the derivative of sin⁡x\sin x.

The product and quotient rules

dydx=x2⋅2e2x+e2x⋅2x=2xe2x(x+1)\frac{\mathrm{d}y}{\mathrm{d}x} = x^2 \cdot 2\mathrm{e}^{2x} + \mathrm{e}^{2x}\cdot 2x = 2x\mathrm{e}^{2x}(x + 1)

dydx=x⋅1x−ln⁡x⋅1x2=1−ln⁡xx2\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x\cdot\frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}

The quotient rule also gives these results, which are worth recognising:

ddx(sec⁡x)=sec⁡xtan⁡xddx(cosec⁡x)=−cosec⁡xcot⁡xddx(cot⁡x)=−cosec⁡2x\frac{\mathrm{d}}{\mathrm{d}x}(\sec x) = \sec x\tan x \qquad \frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{cosec} x) = -\operatorname{cosec} x\cot x \qquad \frac{\mathrm{d}}{\mathrm{d}x}(\cot x) = -\operatorname{cosec}^2 x

Stationary points with these functions

Set the derivative to zero and solve. A positive factor like e−x\mathrm{e}^{-x} can never be zero, so divide it out safely.

dydx=e−x−xe−x=(1−x)e−x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-x} - x\mathrm{e}^{-x} = (1 - x)\mathrm{e}^{-x}. Since e−x>0\mathrm{e}^{-x} > 0, the only solution is x=1x = 1, at the point (1,e−1)\left(1, \mathrm{e}^{-1}\right).

The gradient goes from positive to negative through x=1x = 1, so this is a maximum.

Parametric curves

A curve can be described by writing xx and yy in terms of a third variable, the parameter tt.

dydx=6t22t=3t\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6t^2}{2t} = 3t, which is 3 at t=1t = 1. The point is (2,2)(2, 2), so the tangent is y−2=3(x−2)y - 2 = 3(x - 2), or y=3x−4y = 3x - 4.

Implicit differentiation

When yy is mixed in with xx, differentiate every term with respect to xx. Any term in yy picks up a factor of dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}, and a product of xx and yy needs the product rule.

TermDerivative with respect to xx
y2y^22ydydx2y\dfrac{\mathrm{d}y}{\mathrm{d}x}
xyxyxdydx+yx\dfrac{\mathrm{d}y}{\mathrm{d}x} + y
sin⁡y\sin ycos⁡ydydx\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x}

2x+xdydx+y+2ydydx=0⇒dydx=−2x+yx+2y2x + x\frac{\mathrm{d}y}{\mathrm{d}x} + y + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \quad\Rightarrow\quad \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2x + y}{x + 2y}

At (1,2)(1, 2) the gradient is −45-\dfrac{4}{5}.

The inverse tangent (Paper 3)

ddx(tan⁡−1x)=11+x2ddx(tan⁡−12x)=21+4x2\frac{\mathrm{d}}{\mathrm{d}x}\left(\tan^{-1} x\right) = \frac{1}{1 + x^2} \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(\tan^{-1} 2x\right) = \frac{2}{1 + 4x^2}

Test yourself15 flashcards on this chapter.Flashcards →