Question

Let g(x) be the inverse of an invertible function f(x) which is differentiable for all real x, then g"(f"(x)) equals

• A. $\frac{-f''(x)}{(f'(x))^{3}}$
• B. $\frac{f'(x)f''(x)-(f'(x))^{3}}{f'(x)}$
• C. $\frac{f'(x)f''(x)-(f'(x))^{2}}{f'(x)}$
• D. None of these

Answer 1

Answer: (A)

$\frac{-f''(x)}{(f'(x))^{3}}$

Answer 2

For an invertible function $f(x)$ with inverse $g(x)$, we have

$$g'(y) = \frac{1}{f'(g(y))}\quad \text{where } y = f(x).$$

Differentiating with respect to $y$:

$$g''(y) = -\frac{f''(g(y))}{\big(f'(g(y))\big)^3}.$$

Substitute $y = f(x)$ (so $g(f(x)) = x$):

$$g''(f(x)) = -\frac{f''(x)}{(f'(x))^3}.$$

Thus, the answer is Option (A).