Question
Differentiate:

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

Answer

Let, $\text{y}=\tan(\text{x}^\circ+45^\circ)$
$\Rightarrow \text{y}=\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
$=\sec^2\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}\times\frac{\text{d}}{\text{dx}}(\text{x}+45)\frac{\pi}{180}$
[Using chain rule]
$=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$
So,
$\frac{\text{d}}{\text{dx}}\big\{\tan(\text{x}^\circ+45^\circ)\big\}=\frac{\pi}{180}\sec^2(\text{x}^\circ+45^\circ)$

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