Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 70

Find $f''$ in terms of $g, g'$ and $g''$. If $g$ is a twice differentiable function at $f(x) = xg(x^2)$

By using Chain Rule,


$
\begin{equation}
\begin{aligned}

f(x) =& xg(x^2)
\\
\\
f'(x) =& \frac{d}{dx} (x) \cdot g(x^2) + x \cdot \frac{d}{dx} [g(x^2)]
\\
\\
f'(x) =& 1 \cdot g(x^2) + x \cdot g'(x^2)(2x)
\\
\\
f'(x) =& g(x^2) + 2x^2 g'(x^2)
\\
\\
f''(x) =& \frac{d}{dx} g(x^2) + 2 \left[ \frac{d}{dx} (x^2) \cdot g'(x^2) + x^2 \frac{d}{dx} g'(x^2) \right]
\\
\\
f''(x) =& g'(x^2)(2x) + 2 [ (2x) \cdot g'(x^2) + x^2 g''(x^2) (2x) ]
\\
\\
f''(x) =& 2x g'(x^2) + 2 [ 2x g'(x^2) + 2x^3 g''(x^2) ]
\\
\\
f''(x) =& 2xg'(x^2) + 4 xg' (x^2) + 4x^3 g''(x^2)
\\
\\
f''(x) =& 6x g'(x^2) + 4x^3 g''(x^2)
\end{aligned}
\end{equation}
$