Differentiate the following functions with respect to $x$ :

Solution:

Let $y=e^{a x} \sec (x) \tan (2 x)$

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

$\frac{d y}{d x}=\frac{d}{d x}\left(e^{\operatorname{ax}} \sec x \tan 2 x\right)$

$\frac{d y}{d x}=\frac{d}{d x}\left[e^{\operatorname{ax}} \times(\sec x \tan 2 x)\right]$

We have (uv)' = vu' + uv' (product rule)

$\Rightarrow \frac{d y}{d x}=\sec x \tan 2 x \frac{d}{d x}\left(e^{a x}\right)+e^{a x} \frac{d}{d x}(\sec x \tan 2 x)$

$\Rightarrow \frac{d y}{d x}=\sec x \tan 2 x \frac{d}{d x}\left(e^{a x}\right)+e^{a x} \frac{d}{d x}(\sec x \times \tan 2 x)$

We have (uv)' $=$ vu' $+$ uv' (product rule)

$\Rightarrow \frac{d y}{d x}=\sec x \tan 2 x \frac{d}{d x}\left(e^{a x}\right)+e^{a x} \frac{d}{d x}(\sec x \tan 2 x)$

$\Rightarrow \frac{d y}{d x}=\sec x \tan 2 x \frac{d}{d x}\left(e^{a x}\right)+e^{a x} \frac{d}{d x}(\sec x \times \tan 2 x)$

We will use the product rule once again.

$\Rightarrow \frac{d y}{d x}=\sec x \tan 2 x \frac{d}{d x}\left(e^{a x}\right)+e^{a x}\left[\tan 2 x \frac{d}{d x}(\sec x)+\sec x \frac{d}{d x}(\tan 2 x)\right]$

We know $\frac{d}{d x}\left(e^{x}\right)=e^{x}, \frac{d}{d x}(\sec x)=\sec x \tan x$ and $\frac{d}{d x}(\tan x)=\sec ^{2} x$

$\Rightarrow \frac{d y}{d x}=\sec x \tan 2 x\left[e^{2 x} \frac{d}{d x}(a x)\right]$

$+\mathrm{e}^{\operatorname{ax}}\left[\tan 2 \mathrm{x}(\sec x \tan x)+\sec x\left\{\sec ^{2} 2 x \frac{d}{d x}(2 x)\right\}\right]$

$\Rightarrow \frac{d y}{d x}=a e^{\operatorname{ax}} \sec x \tan 2 x \frac{d}{d x}(x)+e^{\operatorname{ax}}\left[\sec x \tan x \tan 2 x+2 \sec x \sec ^{2} 2 x \frac{d}{d x}(x)\right]$

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$

$\Rightarrow \frac{d y}{d x}=a e^{2 x} \sec x \tan 2 x \times 1+e^{\operatorname{ax}}\left[\sec x \tan x \tan 2 x+2 \sec x \sec ^{2} 2 x \times 1\right]$

$\Rightarrow \frac{d y}{d x}=a e^{a x} \sec x \tan 2 x+e^{a x}\left[\sec x \tan x \tan 2 x+2 \sec x \sec ^{2} 2 x\right]$

$\Rightarrow \frac{d y}{d x}=a e^{\operatorname{ax}} \sec x \tan 2 x+e^{\operatorname{ax}} \sec x\left[\tan x \tan 2 x+2 \sec ^{2} 2 x\right]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\operatorname{ax}} \sec \mathrm{x}\left(\operatorname{atan} 2 \mathrm{x}+\tan \mathrm{x} \tan 2 \mathrm{x}+2 \sec ^{2} 2 \mathrm{x}\right)$

Thus, $\frac{d}{d x}\left(e^{a x} \sec x \tan 2 x\right)=e^{\operatorname{ax}} \sec x\left(\operatorname{atan} 2 x+\tan x \tan 2 x+2 \sec ^{2} 2 x\right)$