Differentiating Special Functions (Cambridge (CIE) IGCSE Additional Maths): Revision Note

• Remember that these rules only work in radians!

a) Find $\mathrm{f}\text{'}\left(x\right)$ for the functions

i. $\mathrm{f}\left(x\right)=\sin x$

ii.$\mathrm{f}\left(x\right)=\cos 2x$

ii.$\mathrm{f}\left(x\right)=3\sin 4x-\cos (2x-3)$

i.

$\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathit{x}$

ii.    Use the chain rule or remember that when $y=\cos \left(ax+b\right)$, then $\frac{\mathrm{d}y}{\mathrm{d}x}=-a\sin \left(ax+b\right)$

$\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{sin}\mathbf{\left(}\mathbf{2}\mathit{x}\mathbf{\right)}$

iii.   Differentiate 'term by term'

$\mathrm{f}\text{'}\left(x\right)=3\left(4\cos 4x\right)-\left(-2\sin \left(2x-3\right)\right)$

$\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{12}\mathbf{cos}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{sin}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}$

b)$\begin{array}{c}\\ \end{array}$ Find the gradient of the tangent to the curve $y=\sin (2x+\frac{\mathrm{\pi }}{6})$ at the point where $x=\frac{\mathrm{\pi }}{8}$.

Gradient of tangent is equal to gradient of curve. To find the gradient, differentiate...

$\frac{\mathrm{d}y}{\mathrm{d}x}=2\cos \left(2x+\frac{\pi }{6}\right)$

... and substitute $x=\frac{\pi }{8}$ into the derivative

$\frac{\mathrm{d}y}{\mathrm{d}x}=2\cos \left(2\left(\frac{\pi }{8}\right)+\frac{\pi }{6}\right)$

Ensure that your calculator is in radians

$=\frac{\sqrt{6}-\sqrt{2}}{2}=0.517638090...$

Answer = $\frac{\sqrt{\mathbf{6}}\mathbf{-}\sqrt{\mathbf{2}}}{\mathbf{2}}$, or 0.518 to 3 significant figures

• Remember to avoid the common mistakes:

  • the derivative of$\ln kx$ with respect to $x$ is$\frac{1}{x}$, NOT$\frac{k}{x}$ !

  • the derivative of ${\mathrm{e}}^{kx}$ with respect to $x$ is $k{\mathrm{e}}^{kx}$, NOT $kx{\mathrm{e}}^{kx-1}$!

A curve has the equation $y={\mathrm{e}}^{-3x}+2\ln x$.

Find the gradient of the curve at the point where $x=2$ giving your answer in the form $a+b{e}^c$, where $a,b$ and $c$ are integers to be found.

Differentiate each term separately. 

$\frac{\mathrm{d}y}{\mathrm{d}x}=-3{\mathrm{e}}^{-3x}+2\left(\frac{1}{x}\right)$

Substitute $x=2$ into the derivative.

$\frac{\mathrm{d}y}{\mathrm{d}x}=-3{\mathrm{e}}^{-3\left(2\right)}+2\left(\frac{1}{2}\right)$ 

Rearrange to the required form.

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& -3{\mathrm{e}}^{-6}+1=1-3{\mathrm{e}}^{-6}\end{array}$

Do not use your calculator to evaluate this as the question asks for the answer given in this form.

$\mathbf{1}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{6}}$