Differentiating exponentials, logarithms and trigonometric functionsAQA A-Level Maths: Revision notes

For a constant $k$: $\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.$ The function $e^{x}$ is its own derivative, and the chain rule supplies the factor $k$. For a base $a>0$, write $a^{kx}=e^{kx\ln a}$, so $\frac{d}{dx}\left(a^{kx}\right)=k\ln a\cdot a^{kx}.$ Example: $y=5\times2^{0.4t}$ gives $\frac{dy}{dt}=5\times0.4\ln2\times2^{0.4t}=2\ln2\times2^{0.4t}$. Sums, differences and constant multiples are differentiated term by term.

$\frac{d}{dx}(\ln x)=\frac1x\quad(x>0).$ Because $\ln(kx)=\ln k+\ln x$, the derivative of $\ln(kx)$ is also $\frac1x$, not $\frac{1}{kx}$. Use laws of logarithms first where they simplify the work: $\ln x^3=3\ln x$ has derivative $\frac3x$. Example: $y=x-2\ln x$ has $\frac{dy}{dx}=1-\frac2x$, which is zero at $x=2$.

With $x$ in radians, and $k$ a constant: $\frac{d}{dx}(\sin kx)=k\cos kx,\qquad\frac{d}{dx}(\cos kx)=-k\sin kx,\qquad\frac{d}{dx}(\tan kx)=k\sec^2kx.$ Example: $y=\tan2x$ has $\frac{dy}{dx}=2\sec^22x$. At $x=\frac{\pi}{8}$, the gradient is $2\sec^2\frac{\pi}{4}=4$ and $y=1$, so the tangent is $y=4x+1-\frac{\pi}{2}$. Combinations such as $y=2\sin3x+\cos2x$ are differentiated term by term: $\frac{dy}{dx}=6\cos3x-2\sin2x$.

The derivative is the limit of the gradient of a chord: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. For $\sin x$, use $\sin(x+h)=\sin x\cos h+\cos x\sin h$: $\frac{\sin(x+h)-\sin x}{h}=\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}.$ As $h\to0$, $\frac{\sin h}{h}\to1$ and $\frac{\cos h-1}{h}\to0$ (small angle results, with $h$ in radians), so the limit is $\cos x$. For $\cos x$, use $\cos(x+h)=\cos x\cos h-\sin x\sin h$. The same limits give $\cos x\times0-\sin x\times1=-\sin x$.

Gradients: substitute the $x$-value into $\frac{dy}{dx}$, then form the tangent $y-y_1=m(x-x_1)$. Stationary points: solve $\frac{dy}{dx}=0$. Example: $y=e^{3x}-4x$ has $\frac{dy}{dx}=3e^{3x}-4=0$ so $e^{3x}=\frac43$ and $x=\frac13\ln\frac43$. Take logarithms to solve equations containing an exponential. Growth models: for $P=P_0a^{kt}$, the rate of growth $\frac{dP}{dt}=k\ln a\times P$ is proportional to $P$.