Differentiate products and quotients of functions, and derive the derivatives of $\tan x$, $\cot x$, $\sec x$ and $\operatorname{cosec} x$.
The derivative of a product is not the product of the derivatives. For example $x \cdot x = x^2$ has derivative $2x$, not $1 \cdot 1$. The correct rule is: for differentiable $u(x)$ and $v(x)$, $$\frac{d}{dx}(uv) = u'v + uv'.$$ In words: derivative of the first times the second, plus the first times the derivative of the second.
Why: $u(x+h)v(x+h) - u(x)v(x) = \big[u(x+h) - u(x)\big]v(x+h) + u(x)\big[v(x+h) - v(x)\big]$. Divide by $h$ and let $h \to 0$ (using $v(x+h) \to v(x)$).
Example: Differentiate $y = x^2\sin x$. Take $u = x^2$, $v = \sin x$: $u' = 2x$, $v' = \cos x$. So $\dfrac{dy}{dx} = 2x\sin x + x^2\cos x$.
Example: Differentiate $y = (3x - 1)\sqrt{x}$. Two ways. By the rule: $u = 3x - 1$, $v = x^{1/2}$, giving $3\sqrt{x} + (3x - 1)\cdot\dfrac{1}{2\sqrt{x}} = \dfrac{6x + 3x - 1}{2\sqrt{x}} = \dfrac{9x - 1}{2\sqrt{x}}$. By expanding first: $y = 3x^{3/2} - x^{1/2}$, so $y' = \tfrac92 x^{1/2} - \tfrac12 x^{-1/2} = \dfrac{9x - 1}{2\sqrt{x}}$. Same answer; expanding is often quicker.
For $y = \dfrac{u}{v}$ with $v \ne 0$, $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.$$ You can remember it as "low d-high minus high d-low, over low squared". It follows from the product rule on $u \cdot v^{-1}$.
Example: Differentiate $y = \dfrac{x^2 + 1}{x - 1}$ ($x \ne 1$). $u = x^2 + 1$, $v = x - 1$, so $u' = 2x$, $v' = 1$. $\dfrac{dy}{dx} = \dfrac{2x(x-1) - (x^2+1)\cdot 1}{(x-1)^2} = \dfrac{x^2 - 2x - 1}{(x-1)^2}$.
Write each as a quotient of $\sin x$ and $\cos x$. For $\tan x = \dfrac{\sin x}{\cos x}$: $$\frac{d}{dx}(\tan x) = \frac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.$$ In the same way (using $\sec x = \dfrac{1}{\cos x}$ and so on):
| function | derivative |
|---|---|
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\operatorname{cosec}^2 x$ |
| $\sec x$ | $\sec x\tan x$ |
| $\operatorname{cosec} x$ | $-\operatorname{cosec} x\cot x$ |
Example: Differentiate $y = \dfrac{\sin x}{1 + \cos x}$. $u = \sin x$, $v = 1 + \cos x$: $u' = \cos x$, $v' = -\sin x$. $\dfrac{dy}{dx} = \dfrac{\cos x(1 + \cos x) + \sin^2 x}{(1 + \cos x)^2} = \dfrac{\cos x + 1}{(1+\cos x)^2} = \dfrac{1}{1 + \cos x}$. (Check: $y = \tan\dfrac{x}{2}$, and its derivative is $\tfrac12\sec^2\dfrac{x}{2} = \dfrac{1}{1+\cos x}$.)
Example: Find the gradient of $y = x\tan x$ at $x = \dfrac{\pi}{4}$. $\dfrac{dy}{dx} = \tan x + x\sec^2 x$. At $x = \dfrac{\pi}{4}$, $\tan x = 1$ and $\sec^2 x = 2$, so the gradient is $1 + \dfrac{\pi}{2}$.
Exam tip: Before reaching for the quotient rule, ask whether the expression can be split or simplified: $\dfrac{x^2 + 3x}{x} = x + 3$ and $\dfrac{1}{\cos x} = \sec x$ are far easier. Paper I answers are usually given in a simplified form, so also cancel common factors (as in the $\dfrac{\sin x}{1+\cos x}$ example) and state the result in terms of the trigonometric functions asked for. A common slip is the order of the numerator $u'v - uv'$: swapping it changes the sign.