Question
Differentiate the following functions with respect to x:
$\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
$\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
✓
Answer
Let, $\text{y}=\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\big)$
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\times\frac{1}{1+(\sqrt{\text{x}})^2}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{1+\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\Big)=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\big)$
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\times\frac{1}{1+(\sqrt{\text{x}})^2}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{1+\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\Big)=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
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