If $\mathrm{y}={\mathrm{e}}^{{\mathrm{x}}^2\mathrm{cosx}}+(\mathrm{cosx}{)}^{\mathrm{x}}$ then find $\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{. }$
 

Given:
$\mathrm{y}={\mathrm{e}}^{{\mathrm{x}}^2\mathrm{cosx}}+(\mathrm{cosx}{)}^{\mathrm{x}}$
Differentiate the equation w.r.t. x,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{x}}^2\cos \mathrm{x}}+\frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x}{)}^{\mathrm{x}}$
Let assume $(\cos \mathrm{x}{)}^{\mathrm{x}}=\mathrm{t}$
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{x}}^2\cos \mathrm{x}}+\frac{\mathrm{dt}}{\mathrm{dx}}\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{{\mathrm{x}}^2\cos \mathrm{x}}\left(2\mathrm{xcos}\mathrm{x}-{\mathrm{x}}^2\sin \mathrm{x}\right)+\frac{\mathrm{dt}}{\mathrm{dx}}(\because \text{ chain rule })......\left(1\right)\end{array}$
Apply ln to the both sides,
$\therefore \ln (\cos \mathrm{x}{)}^{\mathrm{x}}=\ln \mathrm{t}$
Differentiate the equation w.r.t. x,
$\begin{array}{l}\therefore \frac{\mathrm{d}}{\mathrm{dx}}\ln (\cos \mathrm{x}{)}^{\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{dx}}\ln \mathrm{t}\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\mathrm{xln}\cos \mathrm{x}=\frac{\mathrm{d}}{\mathrm{dt}}\ln \mathrm{t}\times \frac{\mathrm{dt}}{\mathrm{dx}}\end{array}$
$\mathrm{x}\frac{1}{\cos \mathrm{x}}(-\sin \mathrm{x})+\ln \cos \mathrm{x}=\frac{1}{\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{dx}}$(∵ 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑟𝑢𝑙𝑒 𝑎𝑛𝑑 𝑐ℎ𝑎𝑖𝑛 𝑟𝑢𝑙𝑒)
$\therefore -\mathrm{xtan} \mathrm{x}+\ln \cos \mathrm{x}=\frac{1}{\mathrm{t}}\frac{dt}{\mathrm{dx}}$
$\begin{array}{l}\therefore \mathrm{t}(-\mathrm{xtan}\mathrm{x}+\ln \cos \mathrm{x})=\frac{\mathrm{dt}}{\mathrm{dx}}\\ \therefore (\cos \mathrm{x}{)}^{\mathrm{x}}\{-\mathrm{xtan}\mathrm{x}+\ln \cos \mathrm{x}\}=\frac{\mathrm{dt}}{\mathrm{dx}}\left(\because (\cos \mathrm{x}{)}^{\mathrm{x}}=\mathrm{t}\right)\end{array}$
Substitute this in (1), it follows
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{{\mathrm{x}}^2\cos \mathrm{x}}\left(2\mathrm{xcos} \mathrm{x}-{\mathrm{x}}^2\sin \mathrm{x}\right)+(\cos \mathrm{x}{)}^{\mathrm{x}}\{-\mathrm{xtan} \mathrm{x}+\ln \cos \mathrm{x}\}$
Which is the answer.