Differentiate the ( ^ - 1 e^ - x ) w.r.t. x. - Vidyadip

Question

Differentiate the $\sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)$ w.r.t. x.

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Answer

Let $y = \sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)$

$\therefore \frac{{dy}}{{dx}} = \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\,\,\left[ {\because \frac{d}{{dx}}\sin f\left( x \right) = \cos f\left( x \right)\frac{d}{{dx}}f\left( x \right)} \right]$

$= \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\frac{d}{{dx}}{e^{ - x}}\,\,\left[ {\because \frac{d}{{dx}}{{\tan }^{ - 1}}f\left( x \right) = \frac{1}{{{{ 1+\left ({f\left( x \right)} \right)}^2}}}\frac{d}{{dx}}f\left( x \right)} \right]$

$= \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{1}{{1 + {e^{ - 2x}}}}{e^{ - x}}\frac{d}{{dx}}\left( { - x} \right)$

$= - \frac{{{e^{ - x}}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}}$

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