Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 89

Show that $\displaystyle \frac{d^2 y}{dx^2} = \frac{d^2 y}{du^2} \left( \frac{du}{dx} \right) ^2 + \frac{dy}{du} \frac{d^2 u}{dx^2} $ having $y = f(u)$ and $u = g(x)$ where $f$ and $g$ are twice differentiable functions.

From the definition of Chain Rule,

$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \left( \frac{du}{dx} \right)$

Thus,



$
\begin{equation}
\begin{aligned}

\frac{d^2y}{dx^2} =& \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{dy}{du} \frac{du}{dx} \right) = \frac{d}{dx} \left( \frac{dy}{du} \right) \frac{du}{dx} + \frac{dy}{du} \frac{d}{dx} \left( \frac{du}{dx} \right)
\\
\\
\frac{d^2y}{dx^2} =& \frac{d}{du} \left( \frac{dy}{du} \right) \left( \frac{du}{dx} \right) \frac{du}{dx} + \frac{dy}{du} \left( \frac{d^2 u}{dx^2} \right)
\\
\\
\frac{d^2y}{dx^2} =& \frac{d^2y}{du^2} \left( \frac{du}{dx} \right) ^2 + \frac{dy}{du} \left( \frac{d^2u}{dx^2} \right)



\end{aligned}
\end{equation}
$