Differentiate the following functions with respect to x :

Let $y=\tan \left(5 x^{\circ}\right)$

First, we will convert the angle from degrees to radians.

We have $1^{\circ}=\left(\frac{\pi}{180}\right)^{c} \Rightarrow 5 \mathrm{x}^{\circ}=5 \mathrm{x} \times \frac{\pi}{180}^{c}$

$\Rightarrow y=\tan \left(5 x \times \frac{\pi}{180}\right)$

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

$\frac{d y}{d x}=\frac{d}{d x}\left[\tan \left(5 x \times \frac{\pi}{180}\right)\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sec ^{2}\left(5 \mathrm{x} \times \frac{\pi}{180}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(5 \mathrm{x} \times \frac{\pi}{180}\right)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\sec ^{2}\left(5 x^{\circ}\right) \frac{\pi}{180} \frac{d}{d x}(5 x)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{180} \sec ^{2}\left(5 \mathrm{x}^{\circ}\right)\left[5 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})\right]$

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

$\Rightarrow \frac{d y}{d x}=\frac{\pi}{180} \sec ^{2}\left(5 x^{\circ}\right)[5]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5 \pi}{180} \sec ^{2}\left(5 \mathrm{x}^{\circ}\right)$

Thus, $\frac{d}{d x}\left(\tan 5 x^{\circ}\right)=\frac{5 \pi}{180} \sec ^{2}\left(5 x^{\circ}\right)$