# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\tan \left(x^{\circ}+45^{\circ}\right)$

Solution:

Let $y=\tan \left(x^{\circ}+45^{\circ}\right)$

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

We have $1^{\circ}=\left(\frac{\pi}{180}\right)^{\mathrm{c}} \Rightarrow(\mathrm{x}+45)^{\circ}=\left[\frac{(\mathrm{x}+45) \pi}{180}\right]^{\mathrm{c}}$

$\Rightarrow y=\tan \left[\frac{(x+45) \pi}{180}\right]$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\tan \left[\frac{(\mathrm{x}+45) \pi}{180}\right]\right\}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sec ^{2}\left[\frac{(\mathrm{x}+45) \pi}{180}\right] \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{(\mathrm{x}+45) \pi}{180}\right]$ [using chain rule]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sec ^{2}\left(\mathrm{x}^{\circ}+45^{\circ}\right) \frac{\pi}{180} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+45)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{180} \sec ^{2}\left(\mathrm{x}^{\circ}+45^{\circ}\right)\left[\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(45)\right]$

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$ and derivative of a constant is 0 .

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{180} \sec ^{2}\left(\mathrm{x}^{\circ}+45^{\circ}\right)[1+0]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{180} \sec ^{2}\left(\mathrm{x}^{\circ}+45^{\circ}\right)$

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left[\tan \left(\mathrm{x}^{\circ}+45^{\circ}\right)\right]=\frac{\pi}{180} \sec ^{2}\left(\mathrm{x}^{\circ}+45^{\circ}\right)$