Answer
${\frac{dy}{dx}}={\frac{2x}{{(1+(e^{x^2})^2})}}\times(e^{(x)^2})$
Work Step by Step
We are given: $y={\tan^{-1}{e^{x^2}}}$
on differentiating both sides:
${\frac{dy}{dx}}={\frac{d({\tan^{-1}{e^{x^2}}})}{dx}}$
on applying the chain rule:
${\frac{dy}{dx}}={\frac{1}{{(1+(e^{x^2})^2})}}\times{\frac{d(e^{(x)^2})}{dx}}$
${\frac{dy}{dx}}={\frac{1}{{(1+(e^{x^2})^2})}}\times(e^{(x)^2}){\frac{d({(x)^2})}{dx}}$
${\frac{dy}{dx}}={\frac{2x}{{(1+(e^{x^2})^2})}}\times(e^{(x)^2})$
The final answer is: ${\frac{dy}{dx}}={\frac{2x}{{(1+(e^{x^2})^2})}}\times(e^{(x)^2})$
