Could someone please help me find the derivative of the inverse of $f$ at $0$?

Question

The problem is:

for $\displaystyle f(x)= \int_0^{\ln x} \frac{1}{\sqrt{4+\mathrm{e}^{t}}} \, \mathrm{d}t$, $x > 0$, find $(f^{-1})'(0)$.

I know that I should use the fundamental theorem of calculus, but I'm having a hard time applying it to this problem.

Any help would be greatly appreciated.

Answer 1

Hint: In general,

$$\frac{d}{dx}\int_{a(x)}^{b(x)}c(t)\,dt = c(b(x))b'(x)-c(a(x))a'(x).$$

What are your $a$, $b$ and $c$ here? After getting an expression for $f'$, what can you do to find $(f^{-1})'(0)$ from, say, $f'(0)$? There is a very useful result for this.

Answer 2

$$ y = f(x), \qquad x = f^{-1}(y), \qquad {f^{-1}}' = (0) = \left.\frac{dx}{dy}\right|_{y=0}. $$

$$ y = \int_0^u \frac{1}{\sqrt{4+e^t}}\, dt,\qquad u = \ln x, \qquad $$ $$ \frac{dy}{du} = \frac{1}{\sqrt{4+e^u}} = \frac{1}{\sqrt{4+x}}, \qquad \frac{du}{dx} = \frac 1 x. $$ $$ \frac{dy}{dx} = \frac{1}{\sqrt{4+x}}\cdot \frac 1 x. $$ $$ \frac{dx}{dy} = x\sqrt{4+x}. $$ $$ \left.\frac{dx}{dy}\right|_{y=0} = 1\cdot\sqrt{4+1} $$ because $y=0$ only when $u=1$, and that happens when $x=1$.