# Derivatives of Inverse Functions

Ok, so you studied inverse functions in precalculus. You know these types of functions are useful but can be abstract. You also know implicit differentiation by now. This article demonstrates a fantastic relationship between the derivative of an inverse of a function and its derivative. To understand what I just said, read on.

So how do you find the derivative of an inverse function? And is there a relationship between the derivative of an inverse function and the derivative of the function?

I also discuss the Implicit Function Theorem and demonstrate its importance through examples. I also derive differentiation formulas for exponential, logarithmic, and inverse trigonometric functions.

## Derivatives of Inverse Functions

In this section we state the derivative rules for the natural exponential function and the general exponential function. We also go over several examples of the chain rule and the exponential derivative rules.

Theorem. If $f$ has an interval $I$ as domain and $f'(x)$ exists and is never zero on $I,$ then $f^{-1}$ is differentiable at every point in its domain. The value of $\left(f^{-1} \right)’$ at a point $b$ in the domain of $f^{-1}$ is the reciprocal of the value of $f’$ at $a=f^{-1}(b)$ given by $$\left(f^{-1}\right)'(b)=\frac{1}{f’\left(f^{-1}(b)\right)}.$$

Example. Let $f$ be the function defined by $$f(x)=\frac{2x-3}{7-5x}.$$ Sketch the graph of $f$ and state whether or not the graph of $f$ passes the horizontal line test. If so, find a rule for $f^{-1}$ and then use it to find $\left(f^{-1}\right)'(0).$ Verify the formula $$\left(f^{-1}\right)'(0)=\frac{1}{f’\left(f^{-1}(0)\right)}.$$

Solution. Notice that the graph of $f$ passes the horizontal line test, which implies that $f$ is a one-to-one function and thus must have an inverse function. To find the inverse function of $f$, set $y=f(x).$ Then simply switch the variables $x$ and $y$ and solve for $y.$ The resulting equation will be a rule for the function $f^{-1}.$ To see this, let $$y=\frac{2x-3}{7-5x}$$ and then switching the variables $x$ and $y$: $$x=\frac{2y-3}{7-5y}.$$ Solving for $y$ yields: $$y=\frac{3+7x}{2+5x}.$$ Thus, $$f^{-1}(x)=\frac{3+7x}{2+5x}.$$ Of course since $f$ and $f^{-1}(x)$ are inverse function it must happen that $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ for each $x$ in the domains. Notice that $$f'(x)=-\frac{1}{(7-5 x)^2} \qquad \text{and} \qquad (f^{-1})'(x)=-\frac{1}{(2+5 x)^2}.$$ Also notice that $$(f^{-1})'(0)=-\frac{1}{(2+5 (0))^2}=-\frac{1}{4}=\frac{1}{f’\left(\frac{3}{2}\right)}=\frac{1}{f’\left(f^{-1}(0)\right)}$$ as desired.

Example. Let $\displaystyle f(x)=\frac{5 x}{1-2x}.$ Without finding a rule for $f^{-1}(x)$ determine $\left(f^{-1}\right) ‘(1).$

Solution. Since $\displaystyle f'(x)=\frac{5}{(1-2 x)^2}$ and $\displaystyle 1=\frac{5 x}{1-2x}$ when $x=1/7$ $$\left(f^{-1}\right)'(1)=\frac{1}{f ‘\left(f^{-1}(1)\right)}=\frac{1}{f ‘(1/7)}=\frac{1}{\frac{49}{5}}=\frac{5}{49}.$$ as needed.

## Derivatives of Exponential Functions

Theorem. The derivative of the exponential function $f(x)=b^x$ is $f'(x)=(\ln b) b^x.$ In the special case when $b=e$ we have $f(x)=e^x$ and $f'(x)=e^x.$ Therefore, $$\frac{d}{dx} \left(b^x\right)=(\ln b)b^x \qquad \text{and} \qquad \frac{d}{dx}\left(e^x\right)=e^x.$$

Example. Find the derivative of the function given $y(x)=x^2-3e^x.$

Solution. The derivative is $\frac{dy}{dx}=2x-3e^x.$

Example. Find the derivative of the function given $y(x)=1-2e^{-x^2}.$

Solution. Using the chain rule the derivative is, $$\frac{dy}{dx}=-2e^{-x^2}\frac{d}{dx}\left(-x^2\right)=-2e^{-x^2}(-2x)=4x e^{-x^2}$$ as desired.

Example. What is the slope of the tangent line to $\displaystyle y=\frac{e^{-x}}{1+e^{-x}}$ at $x=0?$

Solution. Using the quotient rule, \begin{align} \frac{dy}{dx}& =\frac{\left(1+e^{-x}\right)\left(-e^{-x}\right)-e^{-x}\left(-e^{-x}\right)}{\left(1+e^{-x}\right)^2} \\ & =\frac{-e^{-x}\left(1+e^{-x}-e^{-x}\right)}{\left(1+e^{-x}\right)^2} =\frac{-e^{-x}}{\left(1+e^{-x}\right)^2}=-\frac{e^x}{\left(1+e^x\right)^2} \end{align} Therefore, the slope of the tangent line to $y$ at $x=0$ is $$\frac{-e^{-(0)}}{\left(1+e^{-(0)}\right)^2}=-\frac{1}{4}.$$ as desired.

## Derivatives of Logarithmic Functions

Theorem. The derivative of the logarithmic function $f(x)=\log_bx$ is $f'(x)=1/(\ln b) x.$ In the special case when $b=e$ we have $f(x)=\ln x$ and $f'(x)=1/x.$ Therefore, $$\frac{d}{dx} \left(\log_bx\right)=\frac{1}{(\ln b)x} \qquad \text{and} \qquad \frac{d}{dx}(\ln x)=\frac{1}{x}.$$

Example. Find the equation of the tangent line to the curve $y=x^{2 }\ln x$ at $x=e^2.$

Solution. The derivative of $y$ is $y’=2x \ln x+x$ and at $\left(e^2,2 e^4\right)$ we have the slope of the tangent line as $$y’\left(e^2\right)=2\left(e^2\right) \ln \left(e^2\right)+\left(e^2\right)=5 e^2.$$ Therefore, the equation of the tangent line is $y-2e^4=5e^2\left(x-e^2\right)$ which simplifies to $y=5 e^2 x-3 e^4.$

Example. For what values of $A$ and $B$ does $y=A x \ln x+B e^x$ satisfy $y”-y=0?$

Solution. We determine, $y’=A \ln x+A+B e^x$ $y”=\frac{A}{x}+B e^x.$ Since $$y”-y=\frac{A}{x}-A x \ln x=0$$ we find that $A=0$ and that $B$ can be any real number.

Example. Find the derivative of the function given by $$y(x)=5^{2x^2}\ln (4x).$$

Solution. Using the product rule and the chain rule, \begin{align} \frac{dy}{dx} & =\frac{d}{dx}\left(5^{2x^2}\right)\ln (4x)+5^{2x^2}\frac{d}{dx}(\ln (4x)) \\ & =\left(5^{2x^2}\ln (5)(4x)\right) \ln (4x)+5^{2x^2}\left(\frac{1}{4x}(4)\right) \end{align} as desired.

Example. Find the derivative of the function given by $y(x)=\ln (4x+9).$

Solution. Using the chain rule, $$\frac{dy}{dx}=\frac{1}{4x+9}\frac{d}{dx}(4x+9)=\frac{4}{4x+9}.$$

Example. Find the derivative of the function given by $$y(x)=\ln \left(\frac{x^3}{x+1}\right).$$

Solution. Here we simplify first to have, $$y(x)=\ln \left(\frac{x^3}{x+1}\right)=\ln \left(x^3\right)-\ln (x+1)=3 \ln (x)-\ln (x+1)$$ and then using derivative rules, \frac{dy}{dx}=\frac{3}{x}-\frac{1}{x+1}=\frac{2 x+3}{x (x+1)}.
as desired.

Example. Find the derivative of the function given by $$y(x)=\log_2\left(\frac{3x+2}{x^2-5}\right){}^{1/4}.$$

Solution. Here we simplify first to have, $$y(x)=\log_2\left(\frac{3x+2}{x^2-5}\right){}^{1/4}=\frac{1}{4}\log_2(3x+2)-\frac{1}{4}\log_2\left(x^2-5\right)$$ and then using derivative rules, $$\frac{dy}{dx}=\frac{1}{4(3x+2)\ln (2)}\frac{d}{dx}(3x+2)-\frac{1}{4\left(x^2-5\right)\ln (2)}\frac{d}{dx}\left(x^2-5\right)$$ $$\qquad =\frac{3}{4(3x+2)\ln (2)}-\frac{2x}{4\left(x^2-5\right)\ln (2)} =\frac{-3 x^2-4 x-15}{4 (3 x+2) \left(x^2-5\right) \ln (2)}$$ as desired.

Example. Find the derivative of the function given $$s(t)=\log_5\left(\frac{t^2+3}{\sqrt{1-t}}\right).$$

Solution. Here we simplify first to have, $$y(x)=\log_5\left(\frac{t^2+3}{\sqrt{1-t}}\right)=\log_5\left(t^2+3\right)-\frac{1}{2}\log_5(1-t)\$$ and then using derivative rules, $$\frac{dy}{dx}=\frac{2t}{\left(t^2+3\right)\ln (5)}-\frac{(-1)}{2\ln (5)(1-t)}=\frac{3 t^2-4 t-3}{2 (t-1) \left(t^2+3\right) \log (5)}$$ as desired.

## Derivatives of Inverse Trigonometric Functions

Theorem. The inverse trigonometric functions: arcsine, arccosine, arctangent, arccotangent, arccosecant, and arcsecant are all differentiable functions on their domain and their derivative functions are: \begin{align} & \frac{d}{d x}\sin ^{-1} x=\frac{1}{\sqrt{1-x^2}} & & \frac{d}{d x}\cos ^{-1} x=\, -\frac{1}{\sqrt{1-x^2}} \\[.25cm] & \frac{d}{d x}\tan ^{-1}x=\frac{1}{1+x^2} & & \frac{d}{d x}\cot ^{-1}x =\, -\frac{1}{1+x^2} \\[.25cm] & \frac{d}{d x}\sec ^{-1}x=\frac{1}{|x|}\frac{1}{\sqrt{x^2-1}} & & \frac{d}{d x}\csc ^{-1}x=\, -\frac{1}{|x|}\frac{1}{\sqrt{x^2-1}} \end{align}

Example. Find the derivative of the function $y=\sin ^{-1}x\text{ }\cos ^{-1}x.$

Solution. Using the product rule and the derivative formulas for arcsine and arccosine we determine: \begin{align} y’ & =\left(\frac{1}{\sqrt{1-x^2}}\right)\left(\cos ^{-1}x\right)-\left(\frac{1}{\sqrt{1-x^2}}\right)\left(\sin ^{-1}x\right) \\ & =\frac{\cos ^{-1}x-\sin^{-1}x}{\sqrt{1-x^2}} \end{align}
as desired.

Example. Find the derivative of the function $y=\tan ^{-1}x\text{ }\cot ^{-1}x$

Solution. Using the product rule and the derivative formulas for arctangent and arccotangent we determine: \begin{align} y’ & =\left(\frac{1}{1+x^2}\right)\cot ^{-1}x-\left(\frac{1}{1+x^2}\right)\tan ^{-1}x \\ & =\frac{\cot ^{-1}x-\tan ^{-1}x}{1+x^2} \end{align}
as desired.

Example. Find the derivative of the function $y=\tan ^{-1}x-x \sec ^{-1}x$

Solution. Using the product rule and the derivative formulas for arctangent and arcsecant we determine: \begin{align} y’ =\left(\frac{1}{1+x^2}\right)-\sec ^{-1}x-x\left(\frac{1}{|x|}\frac{1}{\sqrt{x^2-1}}\right) \end{align} as desired.

Recall, the Cofunction Theorem from trigonometry: if $\alpha =\cos ^{-1}x$ and $\beta =\sin ^{-1}x$ then $\cos \alpha =\sin \beta$ if and only if $\alpha +\beta =\pi /2.$

Example. Find the derivative of the function $\displaystyle y=\frac{\sin ^{-1}x}{\cos ^{-1}x}.$

Solution. Using the quotient rule, the derivative formulas for arcsine and arccosine and some trigonometric identities, \begin{align} y’&=\frac{\left(\cos ^{-1}x\right)\left(\frac{1}{\sqrt{1-x^2}}\right)-\left(\sin ^{-1} x \right) \left(\frac{-1}{\sqrt{1-x^2}} \right)}{\left(\cos^{-1}x\right)^2} \\ & \qquad = \frac{\cos ^{-1}x+\sin ^{-1}x}{\left(\cos ^{-1}x\right)^2\sqrt{1-x^2}} = \frac{\pi }{2 \sqrt{1-x^2} \left(\cos ^{-1}x\right)^2} \end{align} as desired.

## Exercises on Derivatives of Inverse Functions

Exercise. Find the derivative $\displaystyle \frac{dy}{dx}$ given each of the following.

$(1)\quad y=x^{-3/5}$

$(2)\quad y=7\sqrt{x+6}$

$(3)\quad y=(1-6x)^{2/3}$

$(4)\quad y=\sqrt[3]{x^2}$

$(5)\quad y=\cos (1-6x)^{2/3}$

$(6)\quad y=\sqrt[3]{1+\cos (2x)}$

Exercise. Use implicit differentiation to find $\displaystyle \frac{dy}{dx}$ given each of the following.

$(1)\quad \displaystyle x^3+y^3=18x y$

$(2)\quad \displaystyle x^2(x-y)^2=x^2-y^2$

$(3)\quad \displaystyle x^2=\frac{x-y}{x+y}$

$(4)\quad \displaystyle e^{2x}=\sin (x+3y)$

$(5)\quad \displaystyle e^{x^2y}=2x+2y$

$(6)\quad \displaystyle \sin (x y)=\frac{1}{2}$

Exercise. Use implicit differentiation to find $\displaystyle \frac{dy}{dx}$ at $(1,1)$ for each of the following.

$(1)\quad \displaystyle (x+y)^3=x^3+y^3$

$(2)\quad \displaystyle y^2\left(x^2+y^2\right)=2x^2$

$(3)\quad \displaystyle x\sqrt{1+y}+y\sqrt{1+2x}=2x$

$(4)\quad \displaystyle x^2=\frac{y^2}{y^2-1}$

Exercise. Use implicit differentiation to find $\displaystyle \frac{dy}{dx}$ and then find $\displaystyle \frac{ d^2y}{dx^2}$ given each of the following.

$(1)\quad \displaystyle x^{2/3}+y^{2/3}=1$

$(2)\quad \displaystyle 2\sqrt{y}=x-y.$

Exercise. Verify that the given point is on the given curve and find equations for the lines that are tangent and normal to the curve at this point.

$(1)\quad x^2+x y-y^2=1$ at $(2,3)$

$(2)\quad y^2-2x-4y-1=0$ at $(-2,1)$

$(3)\quad 2x y+\pi \sin y=2\pi$ at $(1,\pi /2)$

$(4)\quad x^2\cos ^2y-\sin y=0$ at $(0,\pi )$

Exercise. Find points on the curve $x^2+ x y+y^2=7$ (a) where the tangent is parallel to the $x$-axis and (b) where the tangent is parallel to the $y$-axis. In the latter case, $\frac{dy}{dx}$ is not defined, but $\frac{dx}{dy}$ is? What value does $\frac{dx}{dy}$ have at these points?

Exercise. Find two points on the curve whose equation is $$x^2-3x y+2y^2=-2,$$ where the tangent line is vertical.

Exercise. Use logarithmic differentiation to find $\frac{dy}{dx}$ for each of the following.

$(1)\quad \displaystyle y=x^{x^3-2x+7}$

$(2)\quad \displaystyle y=x^{\cos x}$

$(3)\quad \displaystyle y=\frac{\sin x}{x^2-3x} e^{x^2}$

$(4)\quad \displaystyle y=(3x-1)^{3x-1}$

$(5)\quad \displaystyle y=\frac{e^{-3x}(x-2)}{\left(x^2+3\right)^2\sqrt{2x-1}}$

$(6)\quad \displaystyle y=x^{\tan x}+x^{\cot x}$

Exercise. Show that the sum of the $x$- and $y$-intercepts of any tangent line to the curve $$\sqrt{x}+\sqrt{y}=\sqrt{c}$$ is equal to $c.$

Exercise. (a) The equations $x^2-2 t x+2t^2=4$ and $2y^3-3t^2=4$ define $x$ and $y$ implicitly as differentiable functions $x=f(t)$ and $y=g(t),$ find the slope of the curve $x=f(t),$ $y=g(t)$ at $t=2.$ (b) The equations $x \sin t+2x=t$ and $t \sin t-2t=y$ define $x$ and $y$ implicitly as differentiable functions $x=f(t)$ and $y=g(t),$ find the slope of the curve $x=f(t),$ $y=g(t)$ at $t=\pi .$

Exercise. Which of the following could be true if $f”(x)=x^{-1/3}?$

$(1)\quad \displaystyle f(x)=\frac{3}{2}x^{2/3}-3$

$(2)\quad \displaystyle f(x)=\frac{9}{10}x^{5/3}-7$

$(3)\quad \displaystyle f”'(x)=\frac{-1}{3}x^{4/3}$

$(4)\quad \displaystyle f(x)=\frac{3}{2}x^{2/3}+6$

Exercise. Use implicit differentiation to show that tangent line at $\left(x_0, y_0\right)$ to any ellipse centered at the origin $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ has the form $$\frac{x_0x}{a^2} + \frac{y_0y}{b^2}=1.$$ Also use implicit differentiation to show that tangent line at $\left(x_0, y_0\right)$ to any hyperbola centered at the origin $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ has the form $\displaystyle \frac{x_0x}{a^2}-\frac{y_0y}{b^2}=1.$

Exercise. Find the derivative $\displaystyle \frac{dy}{dx}$ for each of the following.

$(1)\quad y=\cos ^{-1}\left(x^2\right).$

$(2)\quad y=\sin ^{-1}(1-t).$

$(3)\quad y=\csc ^{-1}\left(x^2+1\right),$ $x>0.$

$(4)\quad y=\sin ^{-1}\left(\frac{3}{t^2}\right).$

$(5)\quad y=\ln \left(\tan ^{-1}x\right).$

$(6)\quad y=\cos ^{-1}\left(e^{-t}\right).$

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.