ln(x+sqrt(x^2+1)) の逆関数の微分

Problem

Let $f(x)=\ln\left(x+\sqrt{x^2+1}\right)$ and $g=f^{-1}$. Find $g'(0)$.

Step 1: Match the inverse input

The input $0$ for the inverse function must be an output of the original function. Since

$$f(0)=\ln\left(0+\sqrt{0^2+1}\right)=\ln(1)=0,$$

the matching value is $x=0$.

Step 2: Differentiate the original function

Next, find the slope of the original function at the matching value. Differentiating

$$f(x)=\ln\left(x+\sqrt{x^2+1}\right)$$

gives

$$f'(x)=\frac{1}{\sqrt{x^2+1}}.$$

Step 3: Evaluate the original slope

Substitute the matching value $x=0$:

$$f'(0)=\frac{1}{\sqrt{0^2+1}}=\frac{1}{1}=1.$$

So the original slope at $0$ is $1$.

Step 4: Use the reciprocal slope rule

For inverse functions,

$$g'(0)=\frac{1}{f'(0)}.$$

Since $f'(0)=1$,

$$g'(0)=1.$$

Thus, the final answer is

$$\boxed{1}$$