Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 46

a.) Show that $\displaystyle g''(x) = \frac{-f''(g(x))}{\left[ f'(g(x))\right]^3}$ Suppose that $f$ is a one-to-one and twice differentiable function with inverse function $g$.
b.) Deduce that if $f$ is decreasing and concave upward, then its inverse function is concave downward.

a.) If $\displaystyle g'(x) = \frac{1}{f'(g(x))}$, then by using Quotient Rule as well as Chain Rule...

$
\begin{equation}
\begin{aligned}
g''(x) = \frac{-f''(g(x)) \cdot g'(x)}{\left[ f'(g(x)) \right]} &= \frac{-f''(g(x)) \cdot \left( \frac{1}{f'(g(x))} \right)}{\left[ f'(g(x)) \right]^2}\\
\\
&= \frac{-f''(g(x))}{\left[ f'(g(x)) \right]^3}
\end{aligned}
\end{equation}
$


b.) Since $f$ is concave upward, $f''(g(x)) > 0$
Thus,
$\displaystyle g''(x) = \frac{f''(g(x))}{\left[ f'(g(x)) \right]^3} < 0$ which shows what $g$ is concave downward.