5.12 Derivatives of Inverse Functions

In this section, we investigate the relationship between the derivative of a function and the derivative of its inverse function.

Let \(f\) be a one-to-one and differentiable function. Because it is differentiable, its graph does not have any corners or any cusps. Because the graph of \(f^{-1}\) is obtained by reflecting the graph of $f$ across the line \(y=x\), the graph of \(f^{-1}\) does not have any corners or any cusps either. We, therefore, expect that \(f^{-1}\) is differentiable wherever the tangent to its graph is not vertical.

Consider a point \(\left(a,f(a)\right)\) on the graph of \(f\) (Figure 1). The equation of the tangent line \(L\) to the graph of \(f\) at \((a,f(a))\) is
\[L:\quad y-f(a)=f'(a)(x-a).\]

 

Figure 1

If the graph of \(f\) and the line \(L\) are reflected through the diagonal line \(y=x\),the graph of \(f^{-1}\) and the tangent line \(L’\) through \((f(a),a)\) are obtained (Figure 2). The equation of \(L’\) is obtained from the equation of \(L\) by interchanging \(x\) and \(y\); that is, \[L’:\quad x-f(a)=f'(a)(y-a)\] or
\[L’:\quad y-a=\frac{1}{f'(a)}(x-f(a)),\] which shows that the slope of $L’$ is $1/f'(a)$. On the other hand, because $L’$ is the tangent line to the graph of $f^{-1}$ at $\left(f(a),a\right)$, the slope of $L’$ is $\left(f^{-1}\right)’\left(f(a)\right)$. Therefore,
\[\left(f^{-1}\right)’\left(f(a)\right)=\frac{1}{f'(a)}.\] Let $b=f(a)$. Then because $a=f^{-1}(b)$, the above equation can be alternatively written as
\[\left(f^{-1}\right)^{\prime}(\underbrace{b}_{f(a)})=\frac{1}{f'(\underbrace{f^{-1}(b)}_{a})}.\]

In general,
\[\bbox[#F2F2F2,5px,border:2px solid black]{(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}}\]
for every \(y\) in the domain of \(f^{-1}\).

 

Figure 2

 

Theorem 1. (The Derivative Rule for Inverses): Assume that \(f\) is a function which is differentiable on the interval \((a,b)\) such that \(f'(x)>0\) (or \(f'(x)<0\)) for all \(x\) in \((a,b).\) Then the inverse function \(x=g(y)\) exists, and we have
\[g'(y)=\frac{1}{f'(x)}=\frac{1}{f'(g(y))},\] or \[g’=\frac{1}{f’\circ g}.\]

  • In the Leibniz notation, \[x=g(y)\Rightarrow g'(y)=\frac{dx}{dy},\] and \[y=f(x)\Rightarrow f'(x)=\frac{dy}{dx}.\] So the above theorem when expressed in the Leibniz notation becomes
    \[\frac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}}.\] Notice that in $dx/dy$ , the independent variable is $y$ , and in $dy/dx$ , the independent variable is $x$ . Additionally notice that if $(x_0 , y_0 )$ is on the graph of $f$ , then $dy/dx$ must be evaluated at $x_0$ and $dx/dy$ must be evaluated at $y_0$, or

    \[\bbox[#F2F2F2,5px,border:2px solid black]{\left(\frac{dx}{dy}\right)_{y_{0}}=\frac{1}{\left(\dfrac{dy}{dx}\right)_{x_{0}}}.}\]

Show the proof

Roughly speaking, because
\[\frac{\Delta x}{\Delta y}=\frac{1}{\dfrac{\Delta y}{\Delta x}}\] and \(\Delta y\to0\) as \(\Delta x\to0\), we have
\[g'(y)=\lim_{\Delta y\to0}\frac{\Delta x}{\Delta y}=\frac{1}{\underset{\Delta x\to0}{\lim}\dfrac{\Delta y}{\Delta x}}=\frac{1}{f'(x)}.\]
However, for a rigorous mathematical proof we need to include more steps and details.

  • The above theorem makes two assertions:
  1. the conditions under which the inverse function, \(g\), is differentiable;
  2. the formula of \(g’\).

 

Show an alternative proof for the above theorem

If it is known that \(g\) is differentiable, we can derive the formula of \(g’\) using implicit differentiation in the following way. Let’s start off with \[x=g(y)\] Then implicitly differentiate \(g(y)=x\) with respect to \(x\): \[\frac{d}{dx}(g(y))=\frac{d}{dx}(x)\] The right hand side is one, and for the left hand side we use the chain rule:
\[\underbrace{\frac{dg}{dy}}_{g'(y)}\underbrace{\frac{dy}{dx}}_{f'(x)}=1.\]
Therefore \[g'(y)=\frac{dx}{dy}=\frac{1}{f'(x)}=\frac{1}{dy/dx}.\]

 

To show how the above formula works, consider \(y=f(x)=x^{2}\). The inverse function is \(x=f^{-1}(y)=\sqrt{y}\). Because \[\frac{dy}{dx}=\frac{df}{dx}=2x,\] we have
\[\begin{align} \frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x}\\ & =\frac{1}{2\sqrt{y}}.\end{align}\]
Thus \((f^{-1})'(y)=\dfrac{1}{2\sqrt{y}}\). Here $y$ is the independent variable, but if we wish, we can denote the independent variable, as usual, by $x$. To do so, we can simply replace $y$ with $x$ on both sides of the equation:

\[(f^{-1})'(x)=\frac{1}{2\sqrt{x}}.\]

Example 1

Let \(f(x)=2x^{3}+3x^{2}+6x+1\). Find \((f^{-1})'(1)\) if it exists.

Solution

The function \(f\) is a polynomial and hence differentiable everywhere. Also because
\[f'(x)=6x^{2}+6x+6=6(x^{2}+x+1)=6\left[\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}\right]>0,\]
its inverse is differentiable everywhere too. It follows from the Derivative Rule for Inverses that
\[(f^{-1})'(1)=\frac{1}{f'(f^{-1}(1))}.\] To find \(f^{-1}(1)\), we need to solve \(f(x)=1\) or \[2x^{3}+3x^{2}+6x+1=1.\] We can see that \(x=0\) is a solution to this equation. Because \(f\) has to be one-to-one to have an inverse, \(x=0\) must be the only solution. So \[(f^{-1})'(1)=\frac{1}{f'(0)}\] and to calculate \(f'(0)\), we just plug \(x=0\) into \(f'(x)\) we already obtained. That is, \[f'(0)=\left.6x^{2}+6x+6\right|_{x=0}=6\] and finally

\[(f^{-1})'(1)=\frac{1}{f'(0)}=\frac{1}{6}.\]

Example 2

Let \(y=f(x)=x^{2}+2x\) for \(x>-1\). Find \(\dfrac{d}{dx}f^{-1}(x)\).

Solution

Method 1: Let’s find \(f^{-1}\) and then differentiate it. To find \(f^{-1}\), we start off with the equation \[y=x^{2}+2x\] and solve it for \(x\): \[x=\frac{-2\pm\sqrt{4-4y}}{2}=-1\pm\sqrt{1+y}.\] This equation gives two values of \(x\) for each \(y\), but we have to choose the one with the \(+\) sign because it is assumed that \(x>-1\). Therefore \[x=f^{-1}(y)=-1+\sqrt{1+y}.\] In the equation \(f^{-1}(y)=-1+\sqrt{1+y}\), \(y\) is a variable that shows the input. We can denote the input with whatever we want, including \(x\). So \[f^{-1}(x)=-1+\sqrt{1+x}.\] Now we can easily find \((f^{-1})'(x)\):
\[\frac{d}{dx}f^{-1}(x)=\frac{1}{2\sqrt{1+x}}.\]
Method 2: Start off with \(y=f(x)\): \[\frac{df}{dx}=\frac{dy}{dx}=2x+2.\] By the Derivative Rule for inverses, we have
\[\begin{align} \frac{df^{-1}}{dy}=\frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x+2}\end{align}\]
Now we need to express \(dx/dy\) in terms of \(y\) because in \(df^{-1}/dy=f'(y)\), the independent variable is \(y\). From \(y=x^{2}+2x\), we solve for \(x\):
\[\begin{align} x & =\frac{-2\pm\sqrt{4+4y}}{2}\\ & =-1\pm\sqrt{1+y}.\end{align}\]
Because \(x>-1\), we set \(x=-1+\sqrt{1+y}\). Therefore
\[\begin{align} \frac{df^{-1}}{dy} & =\frac{1}{2(-1+\sqrt{1+y})+2}\\ & =\frac{1}{2\sqrt{1+y}}.\end{align}\]
Here \(y\) simply shows the input of $f^{-1}$; we can replace it by \(x\) and write:
\[\frac{df^{-1}}{dx}=\left(f^{-1}\right)'(x)=\frac{1}{2\sqrt{1+x}}.\]

Example 3

Suppose \(h(x)=\tan(f^{-1}(x))\) and we know \(f\left(\dfrac{\pi}{4}\right)=1\) and \(f’\left(\dfrac{\pi}{4}\right)=5\). Find \(h'(1)\).

Solution

Let \(u=f^{-1}(x)\). Thus \(h(x)=\tan u\) and using the chain rule, we get
\[\begin{align} h'(x) & =u’\left(\frac{d}{du}\tan u\right)\\ & =u’\cdot(1+\tan^{2}u)\\ & =(f^{-1})'(x)\cdot\left[1+\tan^{2}(f^{-1}(x))\right].\end{align}\]
Because the Derivative Rule for Inverses tells us
\[(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))},\] we have \[h'(x)=\frac{1}{f'(f^{-1}(x))}\left[1+\tan^{2}(f^{-1}(x))\right]\] and
\[h’\left(1\right)=\frac{1}{f'(f^{-1}(1))}\left[1+\tan^{2}(f^{-1}(1))\right].\]
Because \(f(\pi/4)=1\) means \(f^{-1}(1)=\pi/4\), we obtain
\[\begin{align} h'(1) & =\frac{1}{f'(\pi/4)}\left[1+\tan^{2}(\pi/4)\right]\\ & =\frac{1}{5}\left[1+1^{2}\right]\\ & =\frac{2}{5}.\end{align}\]