Tangent line of $(f^{-1})^{'}$, where $f(x) = \int_{0}^{x} 1 + \frac{\sin(\sin(t))}{t} dt$

Question

As the title says, how could I find the tangent line at $x_0 = 0$ of $(f^{-1})^{'}$, where $f(x) = \int_{0}^{x} 1 + \frac{\sin(\sin(t))}{t} dt$?

Thanks in advance!

Answer 1

Sketch. $$ \int_0^x 1 + \frac{\sin\sin t}{t} dt = x + \int_0^x \underbrace{\frac{\sin\sin t}{\sin t}}_{\approx 1}\underbrace{\frac{\sin t}{t} }_{\approx 1}dt $$ for $|x|\ll 1$, which is also valid for $x<0$. So directly from the limit definition, $$\frac{f(x) - f(0)}{x} = 1 + \frac{1}{x} \int_0^x \frac{\sin\sin t}{t} dt \xrightarrow[x\to0]{} 2$$

Thus since $f(x_0) = 0$, the gradient of the tangent line at $x_0$ is $$ (f^{-1})'(0) = \frac1{f'(f^{-1}(0))} = \frac{1}{f'(0)} = \frac{1}{2}$$