Differentiating exponential, logarithmic and trigonometric functionsEdexcel International A Level Maths: Revision notes

The key results (to be known) are: $\frac{d}{dx}\left(e^{kx}\right)=ke^{kx},\qquad\frac{d}{dx}(\ln x)=\frac1x.$ For $\ln kx$, write $\ln kx=\ln k+\ln x$; since $\ln k$ is a constant, $\frac{d}{dx}(\ln kx)=\frac1x$ — the $k$ disappears. Examples: $\frac{d}{dx}(5e^{2x})=10e^{2x}$ and $\frac{d}{dx}(\ln3x)=\frac1x$. The derivative of $e^{kx}$ is a multiple of $e^{kx}$ itself, which is why exponentials model growth and decay.

With $x$ in radians: $\frac{d}{dx}(\sin kx)=k\cos kx,\qquad\frac{d}{dx}(\cos kx)=-k\sin kx,\qquad\frac{d}{dx}(\tan kx)=k\sec^{2}kx.$ The factor $k$ comes from the chain rule. Example: $y=4\sin3x+\cos2x$ gives $\frac{dy}{dx}=12\cos3x-2\sin2x$. Sums and differences are differentiated term by term. The derivative of $\tan kx$ comes from $\tan kx=\frac{\sin kx}{\cos kx}$, with $\sec x=\frac{1}{\cos x}$.

For a positive constant $a$: $\frac{d}{dx}\left(a^{x}\right)=a^{x}\ln a.$ This follows from $a^{x}=e^{x\ln a}$, so the derivative is $\ln a\times e^{x\ln a}$. Example: $\frac{d}{dx}(2^{x})=2^{x}\ln2$, so the gradient at $x=3$ is $8\ln2=5.55$. For $a=e$, $\ln e=1$ gives back $\frac{d}{dx}(e^x)=e^x$.

Differentiate again to find $\frac{d^{2}y}{dx^{2}}$. For $y=5e^{2x}-\ln3x$: $\frac{dy}{dx}=10e^{2x}-x^{-1}$ and $\frac{d^{2}y}{dx^{2}}=20e^{2x}+x^{-2}$. At a stationary point $\frac{dy}{dx}=0$. For trigonometric curves you may need an identity such as $\cos2x=1-2\sin^{2}x$ to form a quadratic in $\sin x$. Worked example: for $y=3\sin2x-2\cos x$, $\frac{dy}{dx}=6\cos2x+2\sin x=0$ gives $6\sin^{2}x-\sin x-3=0$, so $\sin x=\frac{1\pm\sqrt{73}}{12}$. Check the values lie in $[-1,1]$ and in the given interval.