Example 1: Inverse Function Derivative

Differentiate $f^{–1}\left(x\right)=\sqrt[3]{x}$ with respect to x.

Step 1: Simplify and look for algebraic rewrites.

Here the equation is a simplified as we need it, and there are no algebra rewrites because we are not going to apply any derivative rules directly to the inverse function.

You always have to check though.

$f^{–1}\left(x\right)=\sqrt[3]{x}$

Step 2: Make sure you have both the original function, $f\mathrm{(}x\mathrm{)}$, and the inverse function, $f^{–1}\left(x\right)$.

Here we only have the inverse function to start with, but we always need both functions to find the derivative of our inverse.

This means we will have to perform a little algebra to get your original equation.

S1: Swap the position of x and y in the equation.

S2: Solve that new equation for y . In other words, get y alone on one side of the equals. That result is your other equation.

$f^{–1}\left(x\right)=\sqrt[3]{x}$ or rewrite as $y=\sqrt[3]{x}$

S1: $x=\sqrt[3]{y}$

S2: ${\mathrm{(}x\mathrm{)}}^3={\mathrm{(}\sqrt[3]{y}\mathrm{)}}^3$

       $x^3=y$

      $x^3=f\left(x\right)$

We have now both equations.

$f\mathrm{(}x\mathrm{)}=x^3$

$f^{–1}\left(x\right)=\sqrt[3]{x}$

Step 3: Find the derivative of $f\mathrm{(}x\mathrm{)}$

The derivative here requires us to apply the power rule.

Bring the power down.

Subtract 1 from the power.

$f\mathrm{(}x\mathrm{)}=x^3$

$f\prime \left(x\right)= {3x}^2$

Step 4: Construct the $f\prime \mathrm{(}{f}^{–1}\left(x\right)\mathrm{)}$ piece of the inverse derivative rule.

This is a composite function which means we are taking one equation and plugging it into the other equation.

In this derivative rule we will always be plugging the inverse equation, $f^{–1}\left(x\right)$, into the derivative equation, $f\prime \left(x\right)$, everywhere you see an $x$  in the equation.

$f^{–1}\left(x\right)=\sqrt[3]{x}$                 $f\prime \left(x\right)= {3x}^2$

$f\prime \mathrm{(}{f}^{–1}\left(x\right)\mathrm{)}={3\left(\sqrt[3]{x}\right)}^2$

Step 5: Bring the pieces together following the inverse derivative rule.

$\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{f\prime \mathrm{(}{f}^{–1}\left(x\right)\mathrm{)}}$

This really just requires us to take what we found in Step 4, and put it under 1 in the rule.

$\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{f\prime \mathrm{(}{f}^{–1}\left(x\right)\mathrm{)}}$

$\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{{3\left(\sqrt[3]{x}\right)}^2}$

Final Result:

The derivative of$f^{–1}\left(x\right)=\sqrt[3]{x}$   is $\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{{3\left(\sqrt[3]{x}\right)}^2}$.

Most of the time you will want to leave the derivative as it is and not try to expand it or simplify it.

My rule of thumb is if someone wants the derivative give them this unless they ask you to simplify further.

If you need to do more work, like finding a second derivative, you would want to consider expanding and simplifying to make your life easier.

Meaning:

–          The equation for finding the slope of any tangent line at any x-value of $f^{–1}\left(x\right)=\sqrt[3]{x}$ is $\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{{3\left(\sqrt[3]{x}\right)}^2}$.

–          The instantaneous rate of change for every x-value of $f^{–1}\left(x\right)=\sqrt[3]{x}$, is found by using the derivative equation, $\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{{3\left(\sqrt[3]{x}\right)}^2}$.

–          The slope of $f^{–1}\left(x\right)=\sqrt[3]{x}$at any single x-value can be found by plugging it into the derivative, $\mathrm{(}{f}^{–1}\mathrm{)}\prime \left(x\right)=\frac{1}{{3\left(\sqrt[3]{x}\right)}^2}$.