Differentiating standard functionsEdexcel A-Level Maths: Revision notes

The exponential function is its own derivative, and the chain rule brings down a constant factor: $\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.$ For any base $a>0$, $a^{kx}$ is differentiated using $a^{kx}=e^{kx\ln a}$, which gives $\frac{d}{dx}\left(a^{kx}\right)=k\,a^{kx}\ln a.$ Example: $\frac{d}{dx}\left(5^{2x}\right)=2\ln5\times5^{2x}$, and $\frac{d}{dx}\left(4e^{-3x}\right)=-12e^{-3x}$. The gradient of $y=e^{kx}$ at any point is $k$ times the $y$-value, which is why exponential growth and decay are modelled with it.

$\frac{d}{dx}(\ln x)=\frac1x,\qquad x>0.$ Because $\ln x$ is the inverse of $e^{x}$, the gradient of $y=\ln x$ at $x$ is $\frac1x$. Constant multiples and sums work in the usual way: $\frac{d}{dx}(3\ln x)=\frac3x$. Note that $\frac{d}{dx}\ln(kx)=\frac1x$ too, since $\ln(kx)=\ln k+\ln x$ and $\ln k$ is a constant.

With $x$ in radians: $\frac{d}{dx}(\sin kx)=k\cos kx,\quad\frac{d}{dx}(\cos kx)=-k\sin kx,\quad\frac{d}{dx}(\tan kx)=k\sec^{2}kx.$ Example: $\frac{d}{dx}(\sin2x)=2\cos2x$; $\frac{d}{dx}(-\cos2x)=2\sin2x$; $\frac{d}{dx}(\tan3x)=3\sec^{2}3x$. These results only hold in radians. If $x$ is in degrees an extra factor of $\frac{\pi}{180}$ appears.

Differentiate term by term, using the standard results above, and keep constant multiples outside. Example: $y=e^{3x}+\sin2x$ gives $\frac{dy}{dx}=3e^{3x}+2\cos2x$. At $x=0$ the gradient is $3+2=5$, and since $y=1$ the tangent is $y=5x+1$. Example: $y=3\ln x-\cos2x$ gives $\frac{dy}{dx}=\frac3x+2\sin2x$, which is positive for $0<x\le\frac\pi2$, so the curve is increasing there.

The gradient of the chord from $x$ to $x+h$ is $\frac{f(x+h)-f(x)}{h}$, and the derivative is its limit as $h\to0$. For $\sin x$, the addition formula $\sin(x+h)=\sin x\cos h+\cos x\sin h$ gives $\frac{\sin(x+h)-\sin x}{h}=\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right).$ For small $h$ (radians), $\sin h\approx h$ and $\cos h\approx1-\frac{h^2}{2}$, so $\frac{\sin h}{h}\to1$ and $\frac{\cos h-1}{h}\approx-\frac h2\to0$. Therefore $\frac{d}{dx}(\sin x)=\cos x$. The same method with $\cos(x+h)=\cos x\cos h-\sin x\sin h$ gives $\frac{d}{dx}(\cos x)=-\sin x$.