Differentiate the following functions with respect to x :

Solution:

Let $y=x \sin (2 x)+5^{x}+k^{k}+\left(\tan ^{2} x\right)^{3}$

On differentiating $y$ with respect to $x$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left[x \sin 2 x+5^{x}+k^{k}+\left(\tan ^{2} x\right)^{3}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x} \sin 2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}\left(5^{\mathrm{x}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{k}^{\mathrm{k}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\tan ^{2} \mathrm{x}\right)^{3}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x} \times \sin 2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}\left(5^{\mathrm{x}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{k}^{\mathrm{k}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{6} \mathrm{x}\right)$

Recall that (uv)' $=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{d y}{d x}=\sin 2 x \frac{d}{d x}(x)+x \frac{d}{d x}(\sin 2 x)+\frac{d}{d x}\left(5^{x}\right)+\frac{d}{d x}\left(k^{k}\right)+\frac{d}{d x}\left(\tan ^{6} x\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{a}^{\mathrm{x}}\right)=\mathrm{a}^{\mathrm{x}} \log \mathrm{a}, \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$ and $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

Also, the derivation of a constant is 0 .

$\Rightarrow \frac{d y}{d x}=\sin 2 x+x \cos 2 x \frac{d}{d x}(2 x)+5^{x} \log 5+0+6 \tan ^{6-1} x \frac{d}{d x}(\tan x)$

$\Rightarrow \frac{d y}{d x}=\sin 2 x+2 x \cos 2 x \frac{d}{d x}(x)+5^{x} \log 5+6 \tan ^{5} x \frac{d}{d x}(\tan x)$

We have $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$ and $\frac{\mathrm{d}}{\mathrm{dx}}(\tan \mathrm{x})=\sec ^{2} \mathrm{x}$

$\Rightarrow \frac{d y}{d x}=\sin 2 x+2 x \cos 2 x \times 1+5^{x} \log 5+6 \tan ^{5} x \times \sec ^{2} x$

$\therefore \frac{d y}{d x}=\sin 2 x+2 x \cos 2 x+5^{x} \log 5+6 \tan ^{5} x \sec ^{2} x$

Thus, $\frac{d}{d x}\left[x \sin 2 x+5^{x}+k^{k}+\left(\tan ^{2} x\right)^{3}\right]=\sin 2 x+2 x \cos 2 x+5^{x} \log 5+$ $6 \tan ^{5} x \sec ^{2} x$