Differentiation Revision Notes — 9709 Paper 3

The standard derivatives

Syllabus requirement · §3.4.1

“

use the derivatives of eˣ, ln x, sin x, cos x, tan x, tan⁻¹x, together with constant multiples, sums, differences and composites … Derivatives of sin⁻¹x and cos⁻¹x are not required.

”

What you bring from Paper 1

In Paper 1 every function was built from powers of $x$ , and two rules were enough:

the power rule, $\dfrac{\mathrm{d}}{\mathrm{d}x}x^n = nx^{n-1}$ , together with constant multiples, sums and differences;
the chain rule for a function of a function: differentiate the outside function, leave the inside alone, then multiply by the derivative of the inside.

In symbols, if $y = \mathrm{f}\big(\mathrm{g}(x)\big)$ then

\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}'\big(\mathrm{g}(x)\big)\times\mathrm{g}'(x)

For example, $y = \left(3x^2+1\right)^5$ has outside $(\ldots)^5$ and inside $3x^2+1$ . The outside gives $5\left(3x^2+1\right)^4$ , the inside gives $6x$ , and multiplying:

\frac{\mathrm{d}y}{\mathrm{d}x} = 5\left(3x^2+1\right)^4\times 6x = 30x\left(3x^2+1\right)^4

Paper 3 adds six new functions to differentiate. The chain rule does not change at all — it simply gets used with these new outside functions.

The standard derivatives

\frac{\mathrm{d}}{\mathrm{d}x}\,\mathrm{e}^x = \mathrm{e}^x

The gradient equals the height

\frac{\mathrm{d}}{\mathrm{d}x}\ln x = \frac{1}{x}

For x > 0

\frac{\mathrm{d}}{\mathrm{d}x}\sin x = \cos x

x in radians

\frac{\mathrm{d}}{\mathrm{d}x}\cos x = -\sin x

Note the minus; x in radians

\frac{\mathrm{d}}{\mathrm{d}x}\tan x = \sec^2 x

sec²x = 1 + tan²x

\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}x = \frac{1}{1+x^2}

sin⁻¹x and cos⁻¹x are not required

What the formula list gives you

All six of these, and also the derivatives of $\sec x$ , $\csc x$ and $\cot x$ , the product rule, the quotient rule and the parametric rule, are printed in the list of formulae you have in the exam. So the exam does not test whether you remember them. It tests whether you can use them: spot which one a function needs, and apply the chain rule correctly around it.

Where the six results come from

You will not be asked to prove these, but seeing why they are true makes them much harder to misuse.

$\mathrm{e}^x$ . The number $\mathrm{e} \approx 2.718$ is chosen for exactly one reason: it is the base for which the curve $y = \mathrm{e}^x$ has gradient equal to its own height at every point. At $(0, 1)$ the gradient is $1$ ; at $(1, \mathrm{e})$ it is $\mathrm{e}$ . So the derivative of $\mathrm{e}^x$ is $\mathrm{e}^x$ .

$\sin x$ and $\cos x$ . Look at the gradient of $y = \sin x$ as you move along it. At $x = 0$ it climbs most steeply, with gradient $1$ ; at $x = \tfrac12\pi$ it is flat, gradient $0$ ; at $x = \pi$ it falls most steeply, gradient $-1$ . The values $1, 0, -1$ are exactly $\cos 0$ , $\cos\tfrac12\pi$ , $\cos\pi$ — the gradient of $\sin x$ traces out $\cos x$ .

Where sin x climbs, cos x is positive; where sin x is flat, cos x is zero; where sin x falls, cos x is negative. The gradient of the top curve is the height of the bottom one.

Radians, always

The gradient of $\sin x$ at the origin is exactly $1$ only when $x$ is in radians. In degrees the curve is stretched out 57 times wider, the gradient shrinks by a factor of $\dfrac{\pi}{180}$ , and every derivative on this page would be wrong.

So every calculus question on Paper 3 works in radians. If a stationary point comes out at $x = \tfrac14\pi$ , leave it as $\tfrac14\pi$ — never convert it to $45^\circ$ .

The flip rule: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 \div \dfrac{\mathrm{d}x}{\mathrm{d}y}$

The other two results, $\ln x$ and $\tan^{-1}x$ , both come from one simple idea.

A gradient is "rise over run": $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ measures how much $y$ changes per unit change in $x$ . If you turn the question round and ask how much $x$ changes per unit change in $y$ , you get the same triangle read the other way — rise and run swap, so the gradient turns upside down:

\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\;\dfrac{\mathrm{d}x}{\mathrm{d}y}\;}

This is useful whenever $x$ is easy to write in terms of $y$ but $y$ is awkward to write in terms of $x$ . You differentiate $x$ with respect to $y$ , then turn the answer upside down.

Using the flip rule: the derivatives of ln x and tan⁻¹x

(a) Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\ln x = \dfrac1x$ .
(b) Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}x = \dfrac{1}{1+x^2}$ .

Show full working

1
(a) Let $y = \ln x$ . Undo the logarithm by taking $\mathrm{e}$ to the power of both sides: $x = \mathrm{e}^y$
We cannot differentiate $\ln x$ yet, but we can differentiate $\mathrm{e}^y$ . Rewriting with $x$ as the subject is what lets us use a result we already have.
2
Differentiate $x$ with respect to $y$ . The derivative of $\mathrm{e}^y$ is itself: $\frac{\mathrm{d}x}{\mathrm{d}y} = \mathrm{e}^y$
3
Flip it: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\mathrm{e}^y}$
4
The answer should be in terms of $x$ . Since $\mathrm{e}^y = x$ (the first line), replace $\mathrm{e}^y$ by $x$ : $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac1x$
Going back to the original relationship to replace y is the standard last step whenever the flip rule is used.
5
(b) Let $y = \tan^{-1}x$ . Take $\tan$ of both sides to undo the inverse: $x = \tan y$
6
Differentiate $x$ with respect to $y$ : $\frac{\mathrm{d}x}{\mathrm{d}y} = \sec^2 y$
7
We need this in terms of $x$ , and we know $\tan y = x$ . The identity $\sec^2 y = 1 + \tan^2 y$ connects the two: $\frac{\mathrm{d}x}{\mathrm{d}y} = 1 + \tan^2 y = 1 + x^2$
Whenever you have $\sec^2$ and know $\tan$ , this identity is the bridge. The same move appears in exam questions of this shape.
8
Flip it: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1+x^2}$

Answer

(a) $\dfrac1x$ (b) $\dfrac{1}{1+x^2}$

The same four moves (make $x$ the subject, differentiate with respect to $y$ , flip, write in terms of $x$ ) handle exam questions such as "given $\tan y = 2x$ , find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ".

y = tan⁻¹x is y = tan x reflected in y = x. The reflection swaps rise and run, so every gradient turns upside down: 2 at P becomes ½ at the mirror point Q.

The last result, $\dfrac{\mathrm{d}}{\mathrm{d}x}\tan x = \sec^2 x$ , follows from writing $\tan x = \dfrac{\sin x}{\cos x}$ and using the quotient rule — this is worked through in the product and quotient rules section below.

Composites: the chain rule with the new functions

Each standard derivative becomes a chain-rule pattern once the $x$ inside it is replaced by an inside function $\mathrm{f}(x)$ . The derivative of the inside, $\mathrm{f}'(x)$ , always appears as an extra factor:

Function	Outside function gives	Derivative
$\mathrm{e}^{\mathrm{f}(x)}$	$\mathrm{e}^{\mathrm{f}(x)}$	$\mathrm{f}'(x)\,\mathrm{e}^{\mathrm{f}(x)}$
$\ln \mathrm{f}(x)$	$\dfrac{1}{\mathrm{f}(x)}$	$\dfrac{\mathrm{f}'(x)}{\mathrm{f}(x)}$
$\sin \mathrm{f}(x)$	$\cos \mathrm{f}(x)$	$\mathrm{f}'(x)\cos \mathrm{f}(x)$
$\cos \mathrm{f}(x)$	$-\sin \mathrm{f}(x)$	$-\mathrm{f}'(x)\sin \mathrm{f}(x)$
$\tan \mathrm{f}(x)$	$\sec^2 \mathrm{f}(x)$	$\mathrm{f}'(x)\sec^2 \mathrm{f}(x)$
$\tan^{-1} \mathrm{f}(x)$	$\dfrac{1}{1+\left(\mathrm{f}(x)\right)^2}$	$\dfrac{\mathrm{f}'(x)}{1+\left(\mathrm{f}(x)\right)^2}$

The ln row is worth knowing in that shape: it runs backwards in the Integration note, where an integral of the form f′(x)/f(x) gives ln f(x).

Sums, differences and constant multiples work exactly as in Paper 1: differentiate term by term, and a number multiplying a term stays in front. For example

\frac{\mathrm{d}}{\mathrm{d}x}\left(3\mathrm{e}^{x} - 4\ln x + 5\sin x\right) = 3\mathrm{e}^{x} - \frac4x + 5\cos x

Two kinds of composite are easy to miss.

A power of a trig function. $\sin^3 x$ means $\left(\sin x\right)^3$ . The outside function is the cube and the inside is $\sin x$ — not the other way round.

A linear inside. When the inside is something like $3x - 1$ , its derivative is just the number $3$ , and that number sits in front of the answer: $\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{3x-1} = 3\mathrm{e}^{3x-1}$ .

The chain rule with each new function

Differentiate with respect to $x$ :
(a) $y = 4\mathrm{e}^{1-3x}$ (b) $y = \ln\left(x^2+5x\right)$ (c) $y = \cos\left(x^2\right)$ (d) $y = \tan^{-1}(3x)$ (e) $y = \sin^3 x$

Show full working

1
(a) The $4$ is a constant multiple, so it simply stays in front. For $\mathrm{e}^{1-3x}$ , the outside is $\mathrm{e}^{(\ldots)}$ and the inside is $1-3x$ .
Name the outside and the inside before differentiating anything — every mistake in this section comes from skipping this.
2
The outside $\mathrm{e}^{(\ldots)}$ differentiates to itself, $\mathrm{e}^{1-3x}$ . The inside $1-3x$ differentiates to $-3$ .
3
Multiply the pieces together: $\frac{\mathrm{d}y}{\mathrm{d}x} = 4\times\mathrm{e}^{1-3x}\times(-3) = -12\mathrm{e}^{1-3x}$
4
(b) Outside is $\ln(\ldots)$ , inside is $x^2+5x$ . The outside gives $\dfrac{1}{x^2+5x}$ ; the inside gives $2x+5$ .
5
Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x^2+5x}\times(2x+5) = \frac{2x+5}{x^2+5x}$
This is the "derivative of the inside over the inside" pattern. A common error is to write only $\dfrac{1}{x^2+5x}$ and forget the $2x+5$ .
6
(c) Outside is $\cos(\ldots)$ , inside is $x^2$ . The outside gives $-\sin\left(x^2\right)$ ; the inside gives $2x$ .
7
Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = -\sin\left(x^2\right)\times 2x = -2x\sin\left(x^2\right)$
The inside $x^2$ stays inside the sine. Writing $\sin 2x$ here would be differentiating the inside twice.
8
(d) Outside is $\tan^{-1}(\ldots)$ , inside is $3x$ . The outside gives $\dfrac{1}{1+(3x)^2} = \dfrac{1}{1+9x^2}$ ; the inside gives $3$ .
Square the whole inside: $(3x)^2 = 9x^2$ , not $3x^2$ .
9
Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1+9x^2}\times 3 = \frac{3}{1+9x^2}$
10
(e) Rewrite $\sin^3 x$ as $\left(\sin x\right)^3$ . Now the outside is $(\ldots)^3$ and the inside is $\sin x$ .
11
The outside gives $3\left(\sin x\right)^2 = 3\sin^2 x$ ; the inside gives $\cos x$ . Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = 3\sin^2 x\cos x$

Answer

(a) $-12\mathrm{e}^{1-3x}$ (b) $\dfrac{2x+5}{x^2+5x}$ (c) $-2x\sin\left(x^2\right)$ (d) $\dfrac{3}{1+9x^2}$ (e) $3\sin^2x\cos x$

Simplify a logarithm before you differentiate it

The laws of logarithms can turn one hard chain-rule step into several easy ones. For example

y = \ln\frac{(2x+1)^3}{x} = 3\ln(2x+1) - \ln x

by the quotient and power laws, so

\frac{\mathrm{d}y}{\mathrm{d}x} = 3\times\frac{2}{2x+1} - \frac1x = \frac{6}{2x+1} - \frac1x

In the same way $\ln x^4 = 4\ln x$ and $\ln\sqrt{x} = \tfrac12\ln x$ . Differentiating without simplifying first gives the same answer, but with far more algebra.

sec, cosec and cot

These three are printed in the formula list too:

\frac{\mathrm{d}}{\mathrm{d}x}\sec x = \sec x\tan x \qquad \frac{\mathrm{d}}{\mathrm{d}x}\csc x = -\csc x\cot x \qquad \frac{\mathrm{d}}{\mathrm{d}x}\cot x = -\csc^2 x

Questions often tell you to get them yourself — "by writing $\sec^3\theta$ as $\dfrac{1}{\cos^3\theta}$ , show that…" — so you need to know the method: write the function as a power of $\sin$ or $\cos$ , then use the chain rule.

Differentiating sec x from its definition

By writing $\sec x$ as $\dfrac{1}{\cos x}$ , show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\sec x = \sec x\tan x$ .

Show full working

1
Write the function as a power: $\sec x = \frac{1}{\cos x} = \left(\cos x\right)^{-1}$
A power of cos x is a chain-rule problem we already know how to do.
2
The outside is $(\ldots)^{-1}$ , which gives $-1(\ldots)^{-2}$ . The inside is $\cos x$ , which gives $-\sin x$ .
3
Multiply: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\cos x\right)^{-1} = -\left(\cos x\right)^{-2}\times\left(-\sin x\right)$
4
The two minus signs cancel: $= \frac{\sin x}{\cos^2 x}$
Losing one of these two minus signs is the usual mistake. The answer must be positive for small positive x, because sec x is increasing there.
5
Split $\cos^2 x$ into $\cos x \times \cos x$ to recognise the two pieces: $\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\times\frac{\sin x}{\cos x} = \sec x\tan x$

Answer

$\dfrac{\mathrm{d}}{\mathrm{d}x}\sec x = \sec x\tan x$

Common mistakes

$\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{2x} = \mathrm{e}^{2x}$
$2\mathrm{e}^{2x}$
The chain rule still applies: the derivative of the power, 2, is a factor.
$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(3x) = \dfrac{1}{3x}$
$\dfrac{3}{3x} = \dfrac{1}{x}$
It is f′(x)/f(x), not 1/f(x). Here the 3s cancel, which can hide the error.
$\dfrac{\mathrm{d}}{\mathrm{d}x}\cos 2x = \sin 2x$
$-2\sin 2x$
Two errors at once: the minus sign, and the factor 2 from the chain rule.
$\dfrac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(2x) = \dfrac{1}{1+4x^2}$
$\dfrac{2}{1+4x^2}$
The chain rule applies to tan⁻¹ as well.

Your turn

Chain-rule practice with the new functions first — every later section assumes you can do these without pausing — then exam questions on sec, on tan⁻¹, and on the flip rule.

1
Differentiate each of the following with respect to $x$ : (a) $y = \mathrm{e}^{\sin x}$ (b) $y = \ln\!\left(x^2+1\right)$ (c) $y = \tan(3x-1)$ (d) $y = 5\tan^{-1}\!\left(\tfrac12x\right)$
Stuck? Show hint
Each one is an outside function of an inside function. Name both, differentiate each, then multiply.
Show solution
1. 1
  (a) Outside $\mathrm{e}^{(\ldots)}$ , which gives $\mathrm{e}^{\sin x}$ . Inside $\sin x$ , which gives $\cos x$ .
2. 2
  Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = \cos x\,\mathrm{e}^{\sin x}$
3. 3
  (b) Outside $\ln(\ldots)$ , which gives $\dfrac{1}{x^2+1}$ . Inside $x^2+1$ , which gives $2x$ .
4. 4
  Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x}{x^2+1}$
5. 5
  (c) Outside $\tan(\ldots)$ , which gives $\sec^2(3x-1)$ . Inside $3x-1$ , which gives $3$ .
6. 6
  Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = 3\sec^2(3x-1)$
  A linear inside always leaves its coefficient in front.
7. 7
  (d) The $5$ stays in front. Outside $\tan^{-1}(\ldots)$ , which gives $\dfrac{1}{1+\left(\tfrac12x\right)^2} = \dfrac{1}{1+\tfrac14x^2}$ . Inside $\tfrac12x$ , which gives $\tfrac12$ .
8. 8
  Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = 5\times\frac{1}{1+\tfrac14x^2}\times\frac12 = \frac{\tfrac52}{1+\tfrac14x^2}$
9. 9
  Clear the fractions by multiplying top and bottom by $4$ : $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{10}{4+x^2}$
  Not required, but a tidy form is much easier to use in a later part of a question.
Answer
(a) $\cos x\,\mathrm{e}^{\sin x}$ (b) $\dfrac{2x}{x^2+1}$ (c) $3\sec^2(3x-1)$ (d) $\dfrac{10}{4+x^2}$
2
Use the laws of logarithms to simplify $y = \ln\left(x^2\sqrt{4x+1}\right)$ , and hence find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ .
Stuck? Show hint
The log of a product is a sum of logs, and a square root is a power of ½.
Show solution
1. 1
  Product law: $y = \ln x^2 + \ln\sqrt{4x+1}$
2. 2
  Power law on each, since $\sqrt{4x+1} = (4x+1)^{1/2}$ : $y = 2\ln x + \tfrac12\ln(4x+1)$
3. 3
  Differentiate term by term. $\dfrac{\mathrm{d}}{\mathrm{d}x}2\ln x = \dfrac2x$ , and $\ln(4x+1)$ gives $\dfrac{4}{4x+1}$ (inside derivative over inside): $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac2x + \frac12\times\frac{4}{4x+1}$
4. 4
  Simplify the second term: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac2x + \frac{2}{4x+1}$
Answer
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac2x + \dfrac{2}{4x+1}$
3
By writing $\csc x$ as $\left(\sin x\right)^{-1}$ , show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\csc x = -\csc x\cot x$ . Hence differentiate $y = \csc 3x$ .
Stuck? Show hint
Same method as sec x above, with sin in place of cos. For csc 3x, the inside 3x adds a factor of 3.
Show solution
1. 1
  The outside $(\ldots)^{-1}$ gives $-(\ldots)^{-2}$ . The inside $\sin x$ gives $\cos x$ . Multiply: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin x\right)^{-1} = -\left(\sin x\right)^{-2}\cos x = -\frac{\cos x}{\sin^2 x}$
  This time there is only one minus sign, so the answer stays negative.
2. 2
  Split $\sin^2 x$ into $\sin x\times\sin x$ : $-\frac{\cos x}{\sin^2 x} = -\frac{1}{\sin x}\times\frac{\cos x}{\sin x} = -\csc x\cot x$
3. 3
  For $\csc 3x$ : the outside $\csc(\ldots)$ gives $-\csc 3x\cot 3x$ and the inside $3x$ gives $3$ : $\frac{\mathrm{d}y}{\mathrm{d}x} = -3\csc 3x\cot 3x$
Answer
$-\csc x\cot x$ , as required; $\dfrac{\mathrm{d}}{\mathrm{d}x}\csc 3x = -3\csc 3x\cot 3x$
49709/32 M/J 2024 Q10(a)2 marks
By writing $y = \sec^3\theta$ as $\dfrac{1}{\cos^3\theta}$ , show that $\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 3\sin\theta\sec^4\theta$ .
Stuck? Show hint
Write the function as (cos θ)⁻³ and use the chain rule, exactly as for sec x above.
Show solution
1. 1
  Write it as a power of $\cos\theta$ : $y = \frac{1}{\cos^3\theta} = \left(\cos\theta\right)^{-3}$
2. 2
  The outside is $(\ldots)^{-3}$ , which gives $-3(\ldots)^{-4}$ . The inside is $\cos\theta$ , which gives $-\sin\theta$ .
3. 3
  Multiply: $\frac{\mathrm{d}y}{\mathrm{d}\theta} = -3\left(\cos\theta\right)^{-4}\times\left(-\sin\theta\right)$
4. 4
  The two minus signs cancel: $\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{3\sin\theta}{\cos^4\theta}$
5. 5
  Write $\dfrac{1}{\cos^4\theta}$ as $\sec^4\theta$ to reach the printed form: $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 3\sin\theta\sec^4\theta$
  In a “show that”, the last line must be exactly the printed result.
Answer
$\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 3\sin\theta\sec^4\theta$ , as required
59709/35 M/J 2025 Q7(a)3 marks
The equation of a curve is $y = \tan^{-1}(4x)$ . Find the exact values of $x$ when the gradient of the curve is $\tfrac14$ .
Stuck? Show hint
The inside function is 4x. It appears twice in the derivative: squared underneath, and differentiated on top.
Show solution
1. 1
  The outside $\tan^{-1}(\ldots)$ gives $\dfrac{1}{1+(4x)^2} = \dfrac{1}{1+16x^2}$ . The inside $4x$ gives $4$ .
  The mark scheme gives nothing beyond the method mark for $\dfrac{1}{1+4x^2}$ : the whole inside, $4x$ , must be squared.
2. 2
  Multiply: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4}{1+16x^2}$
3. 3
  Set the gradient equal to $\tfrac14$ : $\frac{4}{1+16x^2} = \frac14$
4. 4
  Cross-multiply: $16 = 1 + 16x^2$ , so $16x^2 = 15$ and $x^2 = \dfrac{15}{16}$ .
5. 5
  Take both square roots: $x = \pm\frac{\sqrt{15}}{4}$
  Both signs work, because the gradient only depends on x².
Answer
$x = \dfrac{\sqrt{15}}{4}$ or $x = -\dfrac{\sqrt{15}}{4}$
69709/31 M/J 2024 Q10(a)3 marks
Given that $2x = \tan y$ , show that $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{1 + 4x^2}$ .
Stuck? Show hint
This is the flip rule: differentiate both sides with respect to y, then turn dx/dy upside down.
Show solution
1. 1
  Differentiate both sides of $2x = \tan y$ with respect to $y$ : $2\frac{\mathrm{d}x}{\mathrm{d}y} = \sec^2 y$
2. 2
  So $\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{\sec^2 y}{2}$ , and by the flip rule $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{\sec^2 y}$
3. 3
  Use $\sec^2 y = 1 + \tan^2 y$ : $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1 + \tan^2 y}$
4. 4
  Replace $\tan y$ by $2x$ , so $\tan^2 y = (2x)^2 = 4x^2$ : $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1 + 4x^2}$
  Check: this is the derivative of tan⁻¹(2x) by the chain rule, 2 × 1/(1 + (2x)²).
Answer
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{1 + 4x^2}$ , as required

Practise the standard derivativesReal past-paper questions · Derivatives of e^x, ln x, sin x, cos x, tan x, tan^-1(x) and composites

Checking your access…

Can you do all of these?

Differentiate eˣ, ln x, sin x, cos x, tan x and tan⁻¹x, and composites of them, without dropping the chain-rule factor
Work in radians, and leave answers exact when asked
Simplify a logarithm with the log laws before differentiating it
Use dy/dx = 1 ÷ dx/dy when x is easier to write in terms of y
Differentiate sec, cosec and cot by writing them in terms of sin and cos
Name u and v, differentiate each on its own line, then use the product or quotient rule
Put the quotient rule the right way round: bottom × derivative of top first
Factorise the derivative, and divide only by factors that can never be zero, saying why
Find y as well as x for a stationary point, and state whether it is a maximum or a minimum
Reduce a trig equation to one trig function with an identity, and keep only the roots in the interval
Rearrange dy/dx = 0 into a given form, reaching the tan, ln or e line first
Show that a curve has no stationary points by a range argument, stated in words
For a parametric curve, find t first, then the gradient; use the point's x and y in the line equation
Find horizontal tangents from dy/dt = 0 and vertical tangents from dx/dt = 0
Attach dy/dx every time y is differentiated, and use the product rule on terms like xy
Find the missing coordinate before evaluating an implicit gradient
For tangents parallel to an axis or a line, get a relationship between x and y, then substitute into the curve
Use −1 ÷ (tangent gradient) for a normal, and work backwards from a given normal gradient to the tangent gradient
Show a curve has no horizontal tangent, or no stationary point, with a negative discriminant