Answer
$\frac{d^2y}{dx^2} = (\frac{d^2y}{du^2})(\frac{du}{dx})^2+\frac{dy}{du}~\frac{d^2u}{dx^2}$
Work Step by Step
$y = f(u) = f(g(x))$
$\frac{dy}{dx} = f'(g(x))\cdot g'(x)$
$\frac{d^2y}{dx^2} = f''(g(x))\cdot g'(x)\cdot g'(x)+f'(g(x))\cdot g''(x)$
$\frac{d^2y}{dx^2} = f''(u)\cdot \frac{du}{dx}\cdot \frac{du}{dx}+f'(u)\cdot \frac{d^2u}{dx^2}$
$\frac{d^2y}{dx^2} = (\frac{d^2y}{du^2})(\frac{du}{dx})^2+\frac{dy}{du}~\frac{d^2u}{dx^2}$
