9.2: Inverse Functions and Derivatives

If \(f(x)\) and \(g(x)\) are inverses (remember that means \(f(g(x))=g(f(x))=x\)), then we can find out how their derivatives are related by using the chain rule!

\(\left( f(g(x)) \right)'=(x)' \implies f'(g(x)) g'(x)=1\)

Solving for \(g'(x)\) gives us \(\displaystyle{ g'(x)=\frac{1}{f'(g(x))}}\)

Rewriting \(g(x)=f^{-1}(x)\), this tells us that: \(\displaystyle{ \left(f^{-1}\right)'(x)=\frac{1}{f'(f^{-1}(x))}}\)

If we look at a graph, this should make some sense:

Since the roles of \(x\) and \(y\) are switched for \(f\) and \(f^{-1}\), the slope of the tangent line to \(f^{-1}\) (which is the change in \(y\) over the change in \(x\)) at \((b,a)\) is the reciprocal of the slope of the tangent line to \(f\) at \((a,b)\).

Special note: if \(f'(a)=0\), then \((f^{-1})'(b)\) would be undefined.

We can verify this formula for an example function. Consider \(\textcolor{red}{f(x)=2x^3+1}\) with \(f'(x)=6x^2\)

We can find \(f^{-1}\):

\(x=2y^3+1\)
\(x-1=2y^3\)
\(\displaystyle{\frac{x-1}{2}=y^3}\)
\(\displaystyle{ y=\sqrt[3]{\frac{x-1}{2}}}\)

Thus \(\textcolor{blue}{ \displaystyle{ f^{-1}(x)= \sqrt[3]{\frac{x-1}{2}} } }\)

According to the derivative of inverse formula:

\(\displaystyle{ (f^{-1})'(x)=\frac{1}{f'\left( \textcolor{blue}{f^{-1}(x)}\right)}=\frac{1}{f'\left( \textcolor{blue}{\sqrt[3]{\frac{x-1}{2}}}\right) } = \frac{1}{ 6\left( \textcolor{blue}{ \sqrt[3]{ \frac{x-1}{2} } } \right)^2 } =\frac{1}{6} \left(\frac{x-1}{2}\right)^{-\frac{2}{3}} }\)

If we use the chain rule directly on \(f^{-1}(x)\) instead we get:

\(\displaystyle{\left( \sqrt[3]{\frac{x-1}{2}}\right)'= \frac{1}{3} \left( \frac{x-1}{2}\right)^{-\frac{2}{3}}\cdot \left( \frac{x-1}{2}\right)'=\frac{1}{3} \left( \frac{x-1}{2}\right)^{-\frac{2}{3}}\cdot \frac{1}{2} =\frac{1}{6} \left(\frac{x-1}{2}\right)^{-\frac{2}{3}}}\)

These definitely match, as promised. We can see the graphs of these curves below:

Example \(\PageIndex{1}\)

Find the line tangent to \(\textcolor{blue}{ y=g^{-1}(x)}\) at \(x=3\) when \(\textcolor{red}{g(x)=x^3+x+1}\)

Solution

\(g'(x)=3x^2+1\) which is always positive, so \(g(x)\) is increasing everywhere.

That means the graph of \(y=g(x)\) clearly passes the horizontal line test, and \(g^{-1}\) should exist.

( But if we try to solve for it we get stuck immediately at: \(x=y^3+y\) )

To find the tangent line we need \(g^{-1}(3)\) and \((g^{-1})'(3)\).

\(g^{-1}(3)=b\) for whichever \(b\) makes \(g(b)=3\)

\(b^3+b+1=3 \hspace{.2in} \longleftrightarrow \hspace{.2in} b^3+b-2=0 \hspace{.2in} \longleftrightarrow \hspace{.2in} (b-1)(b^2+b+2)=0\)

Therefore we know \(g^{-1}(3)=1\)

Now we can use the derivative of inverse formula:

\(\displaystyle{ (g^{-1})'(3)=\frac{1}{g'\left( g^{-1}(3)\right)}=\frac{1}{g'(1)}=\frac{1}{3\cdot 1^2+1}=\frac{1}{4}}\)

So our tangent line \(y-g^{-1}(3)=\left(g^{-1}\right)'(3)(x-3)\) becomes \(\textcolor{orange}{\displaystyle{y-1=\frac{1}{4}(x-3)}}\)

Once again everything looks right when we sketch the graphs:

Just like in previous sections, you can do this without the explicit form of a component function:

Example \(\PageIndex{2}\)

Let \(f(x)\) be a function with \(f(1)=2, f(2)=4, f(4)=1, f'(1)=3, f'(2)=5,f'(4)=-2\)
If \(g(x)=f(x^2) \cdot f^{-1}(x)\), find the line tangent to \(y=g(x)\) at \(x=2\).

Solution

We are looking for \(y-g(2)=g'(2)(x-2)\)

We see that since \(f(1)=2\), then \(f^{-1}(2)=1\) so

\(g(2)=f(2^2)\cdot f^{-1}(2)=f(4)\cdot f^{-1}(2)=1\cdot 1 =1\)

Then we differentiate to find \(g'(x)\):

\(\displaystyle{ g'(x)=\left( f(x^2)\right)'\left( f^{-1}(x)\right)+\left( f(x^2)\right)\left( f^{-1}(x)\right)'}\)

\(\displaystyle{ =\left( f'(x^2)\cdot (x^2)'\right)\cdot f^{-1}(x) + f(x^2)\cdot \left( f^{-1}\right)'(x)}\)

\(\displaystyle{ = f'(x^2)\cdot 2x \cdot f^{-1}(x) + f(x^2)\cdot \frac{1}{f'\left( f^{-1}(x)\right)}}\)

Plugging in \(x=2\) we get:

\(\displaystyle{ g'(2)= f'(4)\cdot 4 \cdot f^{-1}(2) + f(4)\cdot \frac{1}{f'\left( f^{-1}(2)\right)}}\)

\(\displaystyle{ = -2 \cdot 4 \cdot 1 + 1\cdot \frac{1}{f'\left( 1\right)}=-8+\frac{1}{3}=-\frac{23}{3}}\)

Therefore the tangent line is: \(\textcolor{red}{\displaystyle{y-1=-\frac{23}{3}(x-2)}}\)

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