Answer
It is an increasing funtion, so it has an inverse (it is one-to-one).
$\displaystyle \frac{df^{-1}}{dx}=\frac{3}{5}x^{-2/5}$
Work Step by Step
Take two values of x,
$x_{1} \gt x_{2}\qquad $... raising to odd power is an increasing function
$x_{1}^{5} \gt x_{2}^{5}\qquad $... cube root is an increasing function
$x_{1}^{5/3} \gt x_{2}^{5/3}$
$ f(x_{1}) \gt f(x_{2})$
So f is an increasing funtion.
By Exercise 49, it is one-to-one and has an inverse.
Theorem 1 provides a way to find a formula for $\displaystyle \frac{df^{-1}}{dx}$
$\displaystyle \frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'[f^{-1}(x)]}$
We find $f^{-1}(x)\quad \left[\begin{array}{l}
y=x^{5/3}\\
x=y^{5/3}\\
y^{5}=x^{3}\\
y=x^{3/5}\\
f^{-1}(x)=x^{3/5}
\end{array}\right]\quad $
We find$ f'(x).$
$f'(x)=\displaystyle \frac{5}{3}x^{2/3}$
Calculate $f'[f^{-1}(x)]$
$f'[x^{3/5}]=\displaystyle \frac{5}{3}[x^{3/5}]^{2/3}=\frac{5}{3}x^{2/5}$
Apply the formula:
$\displaystyle \frac{d}{dx}[f^{-1}(x)]=\frac{1}{\frac{5}{3}x^{2/5}}=\frac{3}{5}x^{-2/5}$
