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4. Differentiate the following w.r.t. x: 

\sin ( \tan ^ { -1} e ^{-x })

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Given function is
f(x)=\sin ( \tan ^ { -1} e ^{-x })
Let's take g(x ) = \tan^{-1}e^{-x}
Now, our function reduces to 
f(x) = \sin(g(x))
Now,
f^{'}(x) = g^{'}(x)\cos(g(x))                   -(i)
And
g(x)=\tan^{-1}e^{-x}\\\Rightarrow g^{'}(x) = \frac{d(\tan^{-1}e^{-x})}{dx}.\frac{d(e^{-x})}{dx}= \frac{1}{1+(e^{-x})^2}.-e^{-x} = \frac{-e^{-x}}{1+e^{-2x}} 
Put this value in our equation (i)
f^{'}(x) =\frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})
Therefore, the answer is \frac{-e^{-x}}{1+e^{-2x}}\cos(\tan^{-1}e^{-x})

Posted by

Gautam harsolia

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