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 $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}.$