CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate the following functions with respect to x:
$\tan(\text{x}^\circ+45^\circ)$
✓
Answer
Let, $\text{y}=\tan(\text{x}^\circ+45^\circ)$
$\Rightarrow\text{y}=\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
Differentiating it with respect to x we get,
$\frac{\text{dx}}{\text{dy}}=\frac{\text{d}}{\text{dx}}\tan\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}$
$=\sec^2\Big\{(\text{x}+45)\frac{\pi}{180}\Big\}\times\frac{\text{d}}{\text{dx}}(\text{x}+45)\frac{\pi}{180}$
[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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