Short and Sweet Calculus

3.9 Derivatives of Inverse Functions

Let \(f\) be a one-to-one and differentiable function. Because \(f\) is differentiable, its graph does not have any corners or any cusps. Since 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 3.9). The equation of the tangent line \(L\) to the graph of \(f\) at \((a,f(a))\) is \[L:\quad y-f(a)=f'(a)(x-a).\] 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 3.9). 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 \(y=f^{-1}(x)\) at \(\left(f(a),a\right)\), the slope of \(L’\) is \(\left(f^{-1}\right)^{\prime}\left(f(a)\right)\). Therefore, \[(f^{-1})'(f(a))=\frac{1}{f'(a)}.\] Let \(b=f(a)\). 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, \[\boxed{(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}}\] for every \(y\) in the domain of \(f^{-1}\) (provided \(f'(f^{-1}(y))\neq0\)).

By the definition of the inverse function, we have \[y=f(x)\quad\Longleftrightarrow\quad x=f^{-1}(y).\] Let’s differentiate \(f\) and \(f^{-1}\): \[y=f(x)\Rightarrow f'(x)=\frac{dy}{dx},\] and \[x=f^{-1}(y)\Rightarrow\left(f^{-1}\right)^{\prime}(y)=\frac{dx}{dy}.\]

3.4. The Derivative Rule for Inverses: If \(y=f(x)\) and \(x=f^{-1}(y)\), then \(dy/dx\) and \(dx/dy\) are related by \[\frac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}}.\]

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 \[\boxed{\left(\frac{dx}{dy}\right)_{y_{0}}=\frac{1}{\left(\dfrac{dy}{dx}\right)_{x_{0}}}.}\]

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 \[\begin{aligned} \frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x}\\ & =\frac{1}{2\sqrt{y}}.\end{aligned}\] Thus \((f^{-1})'(y)=\dfrac{1}{2\sqrt{y}}\) . Here \(y\) is the independent variable. Because mathematical formuals are invariant under permutations of the alphabet, we can have \(x\) as the independent variable as usual, simply by replacing \(y\) with \(x\) on both sides of the equation: \[(f^{-1})'(x)=\frac{1}{2\sqrt{x}}.\]

Example 3.19. 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 3.20. 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 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 \[\begin{aligned} \frac{df^{-1}}{dy}=\frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x+2}\end{aligned}\] 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 \(x\): \[\begin{aligned} x & =\frac{-2\pm\sqrt{4+4y}}{2}\\ & =-1\pm\sqrt{1+y}.\end{aligned}\] Because \(x>-1\), we set \(x=-1+\sqrt{1+y}\). Therefore \[\begin{aligned} \frac{df^{-1}}{dy} & =\frac{1}{2(-1+\sqrt{1+y})+2}\\ & =\frac{1}{2\sqrt{1+y}}.\end{aligned}\] 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 3.21. 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 \[\begin{aligned} h'(x) & =u’\left(\frac{d}{du}\tan u\right)\\ & =u’\cdot(1+\tan^{2}u)\\ & =(f^{-1})'(x)\times(1+\tan^{2}(f^{-1}(x)).\end{aligned}\] 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 \[\begin{aligned} h'(1) & =\frac{1}{f'(\pi/4)}(1+\tan^{2}(\pi/4))\\ & =\frac{1}{5}(1+1^{2})\\ & =\frac{2}{5}.\end{aligned}\]

Example 3.22. Given \(\dfrac{d}{dx}\sin x=\cos x\), find \(\dfrac{d}{dx}\arcsin x\).

[The inverse of sine is regularly denoted by \(\arcsin\) or \(\sin^{-1}\).]

Solution

Recall that \[B=\text{arcsin }A\qquad\text{means}\qquad A=\sin B\quad\text{and}\quad-\frac{\pi}{2}\leq B\leq\frac{\pi}{2}.\]

Let \(y=\sin x\). Then \(x=\arcsin y\) and \(-\pi/2\leq x\leq\pi/2\) \[\begin{aligned} \dfrac{d}{dy}\arcsin y & =\frac{1}{\dfrac{d}{dx}\sin x}\\ & =\frac{1}{\cos x}\\ & =\frac{1}{\sqrt{1-\sin^{2}x}}\tag{\small\ensuremath{\sin^{2}x+\cos^{2}x=1\text{ and }\cos x\geq0}\text{ when }\ensuremath{\frac{-\pi}{2}\leq\emph{x}\leq\frac{\pi}{2}}}\\ & =\frac{1}{\sqrt{1-y^{2}}}\tag{\small\ensuremath{\sin x=y}}\end{aligned}\] With a change in notation we can express the above result as \[\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}.\]


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