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Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Derivatives of eˣ, ln x and trigonometric functions
Pure content (sections A–J) · Subtopic 42 of 66

Derivatives of eˣ, ln x and Trigonometric Functions

The derivatives of exe^x, ln⁡x\ln x, sin⁡x\sin x, cos⁡x\cos x and tan⁡x\tan x, why they need radians, and how to combine them with the chain rule for functions like e3xe^{3x}, ln⁡(x2+1)\ln(x^2+1) and sin⁡2x\sin^2x.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Know the derivatives of exe^x, ln⁡x\ln x, sin⁡x\sin x, cos⁡x\cos x and tan⁡x\tan x (and tan⁡−1x\tan^{-1}x for Paper 3).
  • Use the chain rule with them: ef(x)e^{f(x)}, ln⁡f(x)\ln f(x), sin⁡f(x)\sin f(x) and powers of trigonometric functions.
  • Simplify with log laws before differentiating.
  • Find gradients at points, working in radians.
Pause and think

On the graph of y=sin⁡xy=\sin x (in radians), what is the gradient at x=0x=0? At x=π2x=\tfrac{\pi}{2}? At x=πx=\pi? Which familiar function has those values?

Section 2 of 14

Core idea

Start hereA function that is its own slope

At every point on y=exy=e^x, the gradient equals the height. That one property makes exe^x the language of growth and decay. And the sine wave's gradient traces out a cosine wave. These derivatives, together with the chain rule, unlock most of A-level calculus.

The big idea

ddxex=ex\frac{d}{dx}e^x=e^x, ddxln⁡x=1x\frac{d}{dx}\ln x=\frac1x, ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x=\cos x, ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x=-\sin x, ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x=\sec^2x (xx in radians). With the chain rule: ddxef=f′ef\frac{d}{dx}e^{f}=f'e^f, ddxln⁡f=f′f\frac{d}{dx}\ln f=\frac{f'}f.

1. Why these results

ddxln⁡x=1x\frac{d}{dx}\ln x=\frac1x
  1. y=ln⁡x  ⟺  x=eyy=\ln x\iff x=e^y
    Definition of ln.
  2. dxdy=ey=x\frac{dx}{dy}=e^y=x
    Differentiate with respect to yy.
  3. dydx=1dx/dy=1x\frac{dy}{dx}=\frac1{dx/dy}=\frac1x
    Reciprocal of the derivative.
The gradient of sin x traces cos x
123456−1.5−1−0.50.511.50x (radians)ysin xcos xgradient
The short line is the tangent to sin⁡x\sin x. Its gradient is plotted as the lower point, which always lies on cos⁡x\cos x. This only works when xx is in radians.
ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x=\cos x (first principles)
  1. sin⁡(x+h)−sin⁡xh=sin⁡xcos⁡h+cos⁡xsin⁡h−sin⁡xh\frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}
    Compound formula.
  2. =sin⁡x⋅cos⁡h−1h+cos⁡x⋅sin⁡hh=\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}
    Group.
  3. sin⁡hh→1,cos⁡h−1h→0(h→0, radians)\frac{\sin h}h\to1,\quad\frac{\cos h-1}h\to0\quad(h\to0,\text{ radians})
    Small-angle limits (only true in radians).
  4. →cos⁡x\to\cos x
    Combine.
This is why calculus with trig always uses radians.
ddxekx=kekxddxln⁡(kx)=1xddxsin⁡kx=kcos⁡kxddxcos⁡kx=−ksin⁡kx\frac{d}{dx}e^{kx}=ke^{kx}\quad\frac{d}{dx}\ln(kx)=\frac1x\quad\frac{d}{dx}\sin kx=k\cos kx\quad\frac{d}{dx}\cos kx=-k\sin kx

2. With the chain rule

Worked exampleFive derivatives

Differentiate e3x2e^{3x^2}, ln⁡(x2+5)\ln(x^2+5), cos⁡(4x−1)\cos(4x-1), sin⁡3x\sin^3x, tan⁡2x\tan2x.

Goal
Each derivative.
Strategy
Outside × inside derivative.
  1. e3x2→6xe3x2e^{3x^2}\to6xe^{3x^2}
    Inside 3x23x^2, derivative 6x6x.
  2. ln⁡(x2+5)→2xx2+5\ln(x^2+5)\to\frac{2x}{x^2+5}
    f′f\frac{f'}{f}.
  3. cos⁡(4x−1)→−4sin⁡(4x−1)\cos(4x-1)\to-4\sin(4x-1)
    Inside derivative 44.
  4. (sin⁡x)3→3sin⁡2xcos⁡x(\sin x)^3\to3\sin^2x\cos x
    Power outside, sin inside.
  5. tan⁡2x→2sec⁡22x\tan2x\to2\sec^22x
    Inside 2x2x.
AnswerAs shown.
✓ Checkddxln⁡(x2+5)\frac{d}{dx}\ln(x^2+5) at x=0x=0 is 00: the curve has a minimum there ✓.
Worked exampleUse log laws first: y=ln⁡x2x+1y=\ln\dfrac{x^2}{\sqrt{x+1}}

Differentiate.

Goal
dydx\frac{dy}{dx}.
Strategy
Split the log; much easier than the chain rule on a quotient.
  1. y=2ln⁡x−12ln⁡(x+1)y=2\ln x-\tfrac12\ln(x+1)
    Log laws.
  2. dydx=2x−12(x+1)\frac{dy}{dx}=\frac2x-\frac1{2(x+1)}
    Differentiate each.
Answer2x−12(x+1)\dfrac2x-\dfrac1{2(x+1)}.
✓ CheckCombining: 4(x+1)−x2x(x+1)=3x+42x(x+1)\frac{4(x+1)-x}{2x(x+1)}=\frac{3x+4}{2x(x+1)} (alternative form) ✓.
Edge caseOther bases ax=exln⁡aa^x=e^{x\ln a}, so ddxax=axln⁡a\frac{d}{dx}a^x=a^x\ln a.
Common misconceptionDegrees ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x=\cos x is false in degrees. Always radians.
Pause and think

Gradient of y=e2x−ln⁡xy=e^{2x}-\ln x at x=1x=1?

Summary

  • Standard derivatives table.
  • Chain rule for insides; log laws to simplify.
Key words
Natural logarithm
ln⁡\ln, inverse of exe^x.
The gradient of eˣ equals its value
−2−112−112345670xyy = eˣtangent y = ex(1, e), gradient e(0, 1), gradient 1
Since ddxex=ex\dfrac{d}{dx}e^x=e^x, the gradient at every point equals the yy-value. The tangent at (1,e)(1,e) has gradient ee and passes through the origin.
HistoryArea under a hyperbola. In 1647 Grégoire de Saint-Vincent discovered that areas under y=1xy=\dfrac1x behave like logarithms: they turn multiplication into addition. That is the deep reason ddxln⁡x=1x\dfrac{d}{dx}\ln x=\dfrac1x, and why ln⁡\ln is called the natural logarithm.
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