역삼각함수의 미분

First, we restrict the domain of $f$ to force it to be one–to–one. There are many intervals we could choose over which $f$ is one–to–one, but the question asks us to contain $x=0$ and be as large as possible; this leaves us with the following restricted function:

The inverse of a function swaps the role of the input and output; so if the graph of $y=f(x)$ contains the point $(a,b)$, then the graph of $Y=f^{-1}(X)$ contains the point $(b,a)$. That is, the graph of $Y=f^{-1}(X)$ is the graph of $y=f(x)$ with the $x$-coordinates and $y$-coordinates swapped. (So, since $y=f(x)$ crosses the $y$-axis at $y=1$, then $Y=f^{-1}(X)$ crosses the $X$-axis at $X=1$.) This swapping is equivalent to reflecting the curve $y=f(x)$ over the line $y=x$.

Remark: while you're getting accustomed to inverse functions, it is sometimes clearer to consider $y=f(x)$ and $Y=f^{-1}(X)$: using slightly different notations for $x$ (the input of $f$, hence the output of $f^{-1}$) and $X$ (the input of $f^{-1}$, which comes from the output of $f$). However, the convention is to use $x$ for the inputs of both functions, and $y$ as the outputs of both functions, as is written on the graph above.