In this section, we develop the relation 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 by reflecting its graph across the line \(y=x\), the graph of \(f^{-1}\) is obtained, 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 \(y=x\), it shows the graph of \(f^{-1}\) and the tangent line \(L’\) through \((f(a),a)\) (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
\[L’:\quad y-a=\frac{1}{f'(a)}(x-f(a)).\]

Figure 2

Notice that the slope of \(L’\), which is the tangent line to \(y=f^{-1}(x)\) at \(x=f(a)\), is reciprocal of the slope of \(L\), which is tangent line to \(y=f(x)\) at \(x=a\). Also, recall that the slope of the tangent line at a point is equal to the derivative of the corresponding function at that point. That is, \[(f^{-1})'(f(a))=\frac{1}{f'(a)}.\]

Alternatively, we can express the above formula as
\[\bbox[#F2F2F2,5px,border:2px solid black]{(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}}\]
for every \(y\) in the domain of \(f^{-1}\).

Theorem 1. (The Derivative Rule for Inverses) Assume \(f\) 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
\[g'(y)=\frac{1}{f'(x)}=\frac{1}{f'(g(y))}.\]

  • In Leibniz notations, \[x=g(y)\Rightarrow g'(y)=\frac{dx}{dy},\] and \[y=f(x)\Rightarrow f'(x)=\frac{dy}{dx}.\] So the above theorem when expressed in Leibniz notations becomes
    \[\frac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}}.\]

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 \(g’\).

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 \[\frac{dy}{dx}=\frac{df}{dx}=2x.\] Then
\[\begin{align} \frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x}\\ & =\frac{1}{2\sqrt{y}}.\end{align}\]
Thus \((f^{-1})'(y)=\dfrac{1}{2\sqrt{y}}\) , or by changing the dummy variable to \(x\), we get
\[(f^{-1})'(x)=\frac{1}{2\sqrt{x}}.\]

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
\[f'(x)=6x^{2}+6x+6=6(x^{2}+x+1)=6\left[\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}\right]\neq0,\]
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 \(x\): \[x=\frac{-2\pm\sqrt{4^{2}-4y}}{2}=-1\pm\sqrt{1+y}.\] 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 dummy variable that shows the input. We can depict 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 have to 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
\[\begin{align} \frac{df^{-1}}{dy} & =\frac{1}{2(-1+\sqrt{1+y})+2}\\ & =\frac{1}{2\sqrt{1+y}}.\end{align}\]
Because \(y\) is a dummy variable, 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
\[\begin{align} h'(x) & =u’\left(\frac{d}{du}\tan u\right)\\ & =u’\cdot(1+\tan^{2}u)\\ & =(f^{-1})'(x)\times(1+\tan^{2}(f^{-1}(x)).\end{align}\]
But The Derivative Rule for Inverses tells us
\[(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.\] Therefore, \[h'(x)=\frac{1}{f'(f^{-1}(x))}(1+\tan^{2}(f^{-1}(x))\] and
\[h’\left(1\right)=\frac{1}{f'(f^{-1}(1))}(1+\tan^{2}(f^{-1}(1)).\]
Because \(f(\pi/4)=1\) means \(f^{-1}(1)=\pi/4\), we obtain
\[\begin{align} h'(1) & =\frac{1}{f'(\pi/4)}(1+\tan^{2}(\pi/4))\\ & =\frac{1}{5}(1+1^{2})\\ & =\frac{2}{5}.\end{align}\]

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