Mathematics 9709 · Pure Mathematics 3

Differentiating sin, cos and tan

Three derivatives and two warnings: the derivatives are clean, but watch the two traps.
They are only true in radians.
The cosine one has a minus sign. It is the easy mark people drop the most in the paper.

Common mistake

RADIANS, always.

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

is wrong in degrees. If a trig differentiation question is in degrees, convert to radians first. If you do not, every gradient is wrong and you score zero.

The gradient of sin x is cos x: where sin peaks (gradient 0), cos crosses zero.

The three results

$\dfrac{d}{dx}\sin x=\cos x$ , $\dfrac{d}{dx}\cos x=-\sin x$ (mind the minus), $\dfrac{d}{dx}\tan x=\sec^2 x=\dfrac{1}{\cos^2 x}$ .

(Key Points 4.7, 4.8)

Where sec² x comes from — work tan out once

$\tan x=\dfrac{\sin x}{\cos x}$ , so use the quotient rule:

$\dfrac{d}{dx}\tan x=\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$

Since $\cos^2 x+\sin^2 x=1$ , the top becomes $1$ : $\dfrac{1}{\cos^2 x}=\sec^2 x$ . Work it out once, then just memorise it.

The (ax + b) and power forms — chain rule again

$\dfrac{d}{dx}\sin(ax+b)=a\cos(ax+b)$ ; same pattern for cos (keep the minus) and tan. The $a$ is the inside derivative.

(Key Point 4.9)

Powers of trig need the chain rule too. Read $\sin^3 x$ as $(\sin x)^3$ so the inside (and its derivative) is easy to spot:

Function	Differentiates to
$\sin 2x$	$2\cos 2x$
$\cos(3x-1)$	$-3\sin(3x-1)$
$\sin^3 x=(\sin x)^3$	$3\sin^2 x\cos x$
$\cos^2 3x$	$-6\cos 3x\sin 3x$

Common mistake

The classic slip:

\frac{d}{dx}\sin 2x=\cos 2x

, missing the

\times 2

. And never forget the minus out front on

\cos

. Both are just chain-rule tails.

Combining with everything else

Trig derivatives mix freely with the product/quotient rules and exp/log: $x\cos x$ , $\dfrac{\tan x}{x^2}$ , $(3+2\sin x)^5$ , $e^{\sin x}$ , $\ln(\cos x)$ . (textbook Worked Example 4.9)

Worked example

Gradient of $y = x\cos x$ at $x = \tfrac{\pi}{3}$

Product rule: $\dfrac{dy}{dx}=x(-\sin x)+\cos x\cdot 1=\cos x-x\sin x$ .
Substitute $x=\tfrac{\pi}{3}$ ( $\cos\tfrac{\pi}{3}=\tfrac12$ , $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ ): $\tfrac12-\tfrac{\pi}{3}\cdot\tfrac{\sqrt3}{2}$ .

Answer

\dfrac{dy}{dx}=\dfrac12-\dfrac{\pi\sqrt3}{6}\approx-0.407

Worked example

Real question: $y = \sin x \sin 2x$ , stationary point in $0 < x < \tfrac{\pi}{2}$

9709/32 O/N 2022 Q3. It is a product of two trig pieces, solved by the chain rule on $\sin 2x$ plus a double-angle tidy-up.

Product rule ( $\tfrac{d}{dx}\sin 2x=2\cos 2x$ ): $\dfrac{dy}{dx}=\sin x\,(2\cos 2x)+\sin 2x\,\cos x$ .
Write everything in $\sin x$ using $\cos 2x=1-2\sin^2 x$ and $\sin 2x=2\sin x\cos x$ ; it collapses to $\dfrac{dy}{dx}=2\sin x\,(2-3\sin^2 x)$ .
In $(0,\tfrac{\pi}{2})$ , $\sin x\neq0$ , so $\sin^2 x=\tfrac23$ , giving $\sin x=\sqrt{\tfrac23}$ .

Answer

x=\sin^{-1}\!\sqrt{\tfrac23}=0.955\ \text{(3 s.f.)}

The double-angle swap is the step the mark scheme rewards. It turns a two-piece mess into one equation in $\sin x$ . (9709/32 Oct/Nov 2022 Q3)

Your turn— tap to reveal the worked answer (9709-style: gradient of y = e^(sin x) at x = π/2)

Chain rule, inside $\sin x$ with derivative $\cos x$ : $\dfrac{dy}{dx}=\cos x\,e^{\sin x}$ . At $x=\tfrac{\pi}{2}$ , $\cos\tfrac{\pi}{2}=0$ :

Answer

\dfrac{dy}{dx}=0

Only sin, cos, tan are needed. The reciprocal functions (sec, cosec, cot) sit in the advanced layer.

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Differentiating sin, cos and tan

Three derivatives and two warnings: the derivatives are clean, but watch the two traps.
They are only true in radians.
The cosine one has a minus sign. It is the easy mark people drop the most in the paper.

Common mistake

RADIANS, always.

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

is wrong in degrees. If a trig differentiation question is in degrees, convert to radians first. If you do not, every gradient is wrong and you score zero.

The gradient of sin x is cos x: where sin peaks (gradient 0), cos crosses zero.

The three results

$\dfrac{d}{dx}\sin x=\cos x$ , $\dfrac{d}{dx}\cos x=-\sin x$ (mind the minus), $\dfrac{d}{dx}\tan x=\sec^2 x=\dfrac{1}{\cos^2 x}$ .

(Key Points 4.7, 4.8)

Where sec² x comes from — work tan out once

$\tan x=\dfrac{\sin x}{\cos x}$ , so use the quotient rule:

$\dfrac{d}{dx}\tan x=\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$

Since $\cos^2 x+\sin^2 x=1$ , the top becomes $1$ : $\dfrac{1}{\cos^2 x}=\sec^2 x$ . Work it out once, then just memorise it.

The (ax + b) and power forms — chain rule again

$\dfrac{d}{dx}\sin(ax+b)=a\cos(ax+b)$ ; same pattern for cos (keep the minus) and tan. The $a$ is the inside derivative.

(Key Point 4.9)

Powers of trig need the chain rule too. Read $\sin^3 x$ as $(\sin x)^3$ so the inside (and its derivative) is easy to spot:

Function	Differentiates to
$\sin 2x$	$2\cos 2x$
$\cos(3x-1)$	$-3\sin(3x-1)$
$\sin^3 x=(\sin x)^3$	$3\sin^2 x\cos x$
$\cos^2 3x$	$-6\cos 3x\sin 3x$

Common mistake

The classic slip:

\frac{d}{dx}\sin 2x=\cos 2x

, missing the

\times 2

. And never forget the minus out front on

\cos

. Both are just chain-rule tails.

Combining with everything else

Trig derivatives mix freely with the product/quotient rules and exp/log: $x\cos x$ , $\dfrac{\tan x}{x^2}$ , $(3+2\sin x)^5$ , $e^{\sin x}$ , $\ln(\cos x)$ . (textbook Worked Example 4.9)

Worked example

Gradient of $y = x\cos x$ at $x = \tfrac{\pi}{3}$

Product rule: $\dfrac{dy}{dx}=x(-\sin x)+\cos x\cdot 1=\cos x-x\sin x$ .
Substitute $x=\tfrac{\pi}{3}$ ( $\cos\tfrac{\pi}{3}=\tfrac12$ , $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ ): $\tfrac12-\tfrac{\pi}{3}\cdot\tfrac{\sqrt3}{2}$ .

Answer

\dfrac{dy}{dx}=\dfrac12-\dfrac{\pi\sqrt3}{6}\approx-0.407

Worked example

Real question: $y = \sin x \sin 2x$ , stationary point in $0 < x < \tfrac{\pi}{2}$

9709/32 O/N 2022 Q3. It is a product of two trig pieces, solved by the chain rule on $\sin 2x$ plus a double-angle tidy-up.

Product rule ( $\tfrac{d}{dx}\sin 2x=2\cos 2x$ ): $\dfrac{dy}{dx}=\sin x\,(2\cos 2x)+\sin 2x\,\cos x$ .
Write everything in $\sin x$ using $\cos 2x=1-2\sin^2 x$ and $\sin 2x=2\sin x\cos x$ ; it collapses to $\dfrac{dy}{dx}=2\sin x\,(2-3\sin^2 x)$ .
In $(0,\tfrac{\pi}{2})$ , $\sin x\neq0$ , so $\sin^2 x=\tfrac23$ , giving $\sin x=\sqrt{\tfrac23}$ .

Answer

x=\sin^{-1}\!\sqrt{\tfrac23}=0.955\ \text{(3 s.f.)}

The double-angle swap is the step the mark scheme rewards. It turns a two-piece mess into one equation in $\sin x$ . (9709/32 Oct/Nov 2022 Q3)

Your turn— tap to reveal the worked answer (9709-style: gradient of y = e^(sin x) at x = π/2)

Chain rule, inside $\sin x$ with derivative $\cos x$ : $\dfrac{dy}{dx}=\cos x\,e^{\sin x}$ . At $x=\tfrac{\pi}{2}$ , $\cos\tfrac{\pi}{2}=0$ :

Answer

\dfrac{dy}{dx}=0

Only sin, cos, tan are needed. The reciprocal functions (sec, cosec, cot) sit in the advanced layer.