Answer
\[\frac{dy}{dx}=2x\cdot f'\left( g\left( {{x}^{2}} \right) \right)\cdot g'\left( {{x}^{2}} \right)\]
Work Step by Step
\[\begin{align}
& \text{Let }y=f\left( g\left( {{x}^{2}} \right) \right) \\
& \text{Differentiate both sides with respect to }x \\
& \frac{dy}{dx}=\frac{d}{dx}\left[ f\left( g\left( {{x}^{2}} \right) \right) \right] \\
& \text{By the chain rule} \\
& \frac{dy}{dx}=f'\left( g\left( {{x}^{2}} \right) \right)\frac{d}{dx}\left[ g\left( {{x}^{2}} \right) \right] \\
& \frac{dy}{dx}=f'\left( g\left( {{x}^{2}} \right) \right)g'\left( {{x}^{2}} \right)\frac{d}{dx}\left[ {{x}^{2}} \right] \\
& \frac{dy}{dx}=f'\left( g\left( {{x}^{2}} \right) \right)g'\left( {{x}^{2}} \right)\left( 2x \right) \\
& \text{Simplifying} \\
& \frac{dy}{dx}=2x\cdot f'\left( g\left( {{x}^{2}} \right) \right)\cdot g'\left( {{x}^{2}} \right) \\
\end{align}\]
