Papers 2 and 3 (the derivative of \(\tan^{-1} x\) is Paper 3 only). Six new standard derivatives, \(e^{x}\), \(\ln x\), \(\sin x\), \(\cos x\), \(\tan x\) and \(\tan^{-1} x\), combined with the chain rule, let you differentiate almost any function you meet, then find gradients, tangents and stationary points.
\(\dfrac{d}{dx}e^{x} = e^{x}\), \(\dfrac{d}{dx}\ln x = \dfrac{1}{x}\), \(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}\cos x = -\sin x\), \(\dfrac{d}{dx}\tan x = \sec^2 x\), \(\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}\) (x in radians). For a composite, multiply by the derivative of the inside: \(\dfrac{d}{dx}e^{f(x)} = f'(x)e^{f(x)}\) and \(\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}\).
Fig. 1: the gradient of y = ln x is 1/x. At P(2, ln 2) the tangent has gradient ½, and it crosses the y-axis at ln 2 − 1 ≈ −0.307. As x grows, 1/x shrinks, so the curve keeps rising but ever more slowly
Section 01
Exponentials and logarithms.
\(e^{x}\) is the one function that is its own derivative. Its inverse, \(\ln x\), has derivative \(\dfrac{1}{x}\) (Fig. 1). With the chain rule, the inside function's derivative appears as a factor.
\(\dfrac{d}{dx}\ln(4x) = \dfrac{4}{4x} = \dfrac{1}{x}\), NOT \(\dfrac{4}{x}\) and not \(\dfrac{1}{4x}\). (Check: \(\ln 4x = \ln 4 + \ln x\), and \(\ln 4\) is a constant.)
Section 02
Trigonometric functions.
These results are only true when \(x\) is in radians. In calculus questions, assume radians unless told otherwise, and set your calculator to radian mode.
\(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}\cos x = -\sin x\), \(\dfrac{d}{dx}\tan x = \sec^2 x\).
Linear inside: \(\dfrac{d}{dx}\sin(ax + b) = a\cos(ax + b)\), \(\dfrac{d}{dx}\cos(ax + b) = -a\sin(ax + b)\), \(\dfrac{d}{dx}\tan(ax + b) = a\sec^2(ax + b)\).
Powers: \(\sin^3 x = (\sin x)^3\), so use the chain rule with the power outside.
Find the gradient of \(y = \tan 3x\) at \(x = \dfrac{\pi}{12}\).
\(\dfrac{dy}{dx} = 3\sec^2 3x\).
At \(x = \dfrac{\pi}{12}\): \(3x = \dfrac{\pi}{4}\), and \(\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\), so \(\sec^2\dfrac{\pi}{4} = 2\).
Gradient \(= 3 \times 2 = 6\).
Answer: \(6\)
Examiner trap
The minus sign belongs to cosine: \(\dfrac{d}{dx}\cos x = -\sin x\), but \(\dfrac{d}{dx}\sin x = +\cos x\). A memory aid: going round sin, cos, \(-\sin\), \(-\cos\), each step is one derivative.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Differentiate \(e^{5x}\).
Quick check+10 XP
Differentiate \(\ln(4x)\).
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Differentiate \(\cos 3x\).
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(Paper 3 only) Differentiate \(\tan^{-1} x\).
Section 03
The inverse tangent.
Paper 3 only. The derivative of \(\tan^{-1} x\) is a rational function, which is why \(\tan^{-1}\) appears again when you integrate \(\dfrac{1}{x^2 + a^2}\).
\(\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}\).
Chain rule: \(\dfrac{d}{dx}\tan^{-1}(ax) = \dfrac{a}{1 + a^2x^2}\); for example \(\dfrac{d}{dx}\tan^{-1}\dfrac{x}{3} = \dfrac{1/3}{1 + x^2/9} = \dfrac{3}{9 + x^2}\).
Derivatives of \(\sin^{-1} x\) and \(\cos^{-1} x\) are not required.
Worked example Tangent to y = tan⁻¹ x
Find the equation of the tangent to \(y = \tan^{-1} x\) at the point where \(x = 1\).
At \(x = 1\), \(y = \tan^{-1} 1 = \dfrac{\pi}{4}\).
\(\tan^{-1} x\) is the inverse function, not \(\dfrac{1}{\tan x} = \cot x\). Their derivatives are completely different: \(\dfrac{1}{1 + x^2}\) versus \(-\operatorname{cosec}^2 x\).
Section 04
Using the derivatives.
Everything from Pure 1 still applies: the gradient gives tangents and normals, and \(\dfrac{dy}{dx} = 0\) gives stationary points. The equations you solve now involve \(e\), \(\ln\) and trig, so exact answers often contain \(\ln\).
Worked example A stationary point
Find the exact coordinates of the stationary point of \(y = e^{2x} - 6x\) and determine its nature.