5.6 Further differentiation rulesIB Maths: Analysis and Approaches SL: Revision notes

The power rule works for any rational power: if $f(x)=x^{n}$ with $n\in\mathbb{Q}$, then $f'(x)=nx^{n-1}$. Rewrite roots and fractions as powers first: $\sqrt{x}=x^{1/2}$, $\frac{4}{x^{2}}=4x^{-2}$, $\frac{1}{\sqrt[3]{x}}=x^{-1/3}$.

The other standard derivatives (all in the formula booklet) are $\frac{d}{dx}(\sin x)=\cos x,\quad \frac{d}{dx}(\cos x)=-\sin x,\quad \frac{d}{dx}(e^{x})=e^{x},\quad \frac{d}{dx}(\ln x)=\frac{1}{x}.$ Angles must be in radians for the trigonometric results to hold.

Differentiate term by term, and keep constant multiples: $\frac{d}{dx}\big(af(x)+bg(x)\big)=af'(x)+bg'(x)$. For example $\frac{d}{dx}\left(3\sqrt{x}+\frac{4}{x^{2}}-2\ln x\right)=\frac{3}{2\sqrt{x}}-\frac{8}{x^{3}}-\frac{2}{x}.$ There is no such rule for products or quotients: $\frac{d}{dx}(uv)\neq u'v'$.

For a composite function $y=f(g(x))$, let $u=g(x)$. Then $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.$ In words: differentiate the outer function (leaving the inside alone), then multiply by the derivative of the inside.

• $\frac{d}{dx}\sin(3x-1)=3\cos(3x-1)$
• $\frac{d}{dx}e^{x^{2}+2}=2xe^{x^{2}+2}$
• $\frac{d}{dx}(2x+1)^{5}=10(2x+1)^{4}$
• $\frac{d}{dx}\ln(g(x))=\frac{g'(x)}{g(x)}$

If $y=uv$ where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.$ Example: $y=20te^{-0.5t}$ with $u=20t$, $v=e^{-0.5t}$ gives $\frac{dy}{dt}=20e^{-0.5t}-10te^{-0.5t}=10e^{-0.5t}(2-t)$. Factorise the answer where you can: it makes solving $\frac{dy}{dx}=0$ much easier, and an exponential factor is never zero.

If $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}.$ The order in the numerator matters: bottom times derivative of top, minus top times derivative of bottom. Example: $\frac{d}{dx}\left(\frac{\ln x}{x}\right)=\frac{x\cdot\frac1x-\ln x}{x^{2}}=\frac{1-\ln x}{x^{2}}$.

The denominator $v^{2}$ is never negative, so the sign of the derivative is decided by the numerator alone. That is how you find where a quotient is increasing or decreasing.

A derivative such as $C'(t)$ is a rate of change: the units are (units of $C$) per (unit of $t$), e.g. mg L$^{-1}$ per hour. A positive value means the quantity is increasing at that instant; a negative value means it is decreasing. When asked to interpret, say what is changing, whether it is increasing or decreasing, how fast, and at what time.