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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate the following w.r.t.x: $\sin(\tan^{-1}\text{e}^{-\text{x}})$
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Answer

$\text{Let}\ \text{y}=\sin(\tan^{-1}\text{e}^{-\text{x}})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\cos(\tan^{-1}\text{e}^{-\text{x}}).\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{e}^{-\text{x}})= \bigg[\because\frac{\text{d}}{\text{dx}}\sin\text{f(x)}=\cos\text{f(x)}\frac{\text{d}}{\text{dx}}\text{f(x)}\bigg]$
$=\cos(\tan^{-1}\text{e}^{-\text{x}}).\frac{1}{1+(\text{e}^{-\text{x}})^{2}}\frac{\text{d}}{\text{dx}}\text{e}^{-\text{x}}= \bigg[\because\frac{\text{d}}{\text{dx}}\ \tan^{-1} \text{f(x)}=\frac{1}{(\text{f(x)})^{2}}\frac{\text{d}}{\text{dx}}\text{f(x)}\bigg]$
$=\cos(\tan^{-1}\text{e}^{-\text{x}}).\frac{1}{1+\text{e}^{-\text{2x}}}\text{e}^{\text{-x}}\frac{\text{d}}{\text{dx}}({-\text{x}})$
$=\frac{-\text{e}^{-\text{x}}.\cos(\tan^{-1}\text{e}^{-\text{x}})}{1+\text{e}^{-2\text{x}}}$

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