## 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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