# Derivatives of Inverse Functions

Around the time you’re studying exponential and logarithmic differentiation and integration, you’ll probably learn how to get the derivative of an inverse function. This is because some of the derivations of the exponential and log derivatives were a direct result of differentiating inverse functions.

# Reviewing Inverses of Functions

We learned about inverse functions here in the Inverses of Functions section. You get the inverse of a function if you switch the $$x$$ and $$y$$ and solve for the “new $$y$$”. A function has an inverse function if it is one-to-one (or invertible), which means it passes both vertical and horizontal line tests. A function that has an inverse or is one-to-one is strictly monotonic (either increasing or decreasing) for its entire domain.

# Monotonic Functions in an Interval

We can determine if a function is monotonic in an interval (and therefore has an inverse in that interval) if the derivative of that function is either greater than 0 (increasing) or less than 0 (decreasing) for that entire interval.

Let’s first do some problems where we use the derivative to find out if a function has is strictly monotonic (has a strictly increasing or decreasing derivative) on its entire domain:

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 Monotonic Function Problem Solution Strictly monotonic on entire domain?   $$\displaystyle f\left( x \right)=5-x-4{{x}^{3}}$$ $${f}’\left( x \right)=-1-12{{x}^{2}}<0$$   The derivative is negative for all $$x$$, so $$f\left( x \right)$$ is decreasing on $$\left( {-\infty ,\,\infty } \right)$$. Thus $$f\left( x \right)$$ is strictly monotonic and has an inverse. Strictly monotonic on entire domain?   $$\displaystyle f\left( x \right)=5{{x}^{3}}-5x$$ $${f}’\left( x \right)=15{{x}^{2}}-5=5\left( {3{{x}^{2}}-1} \right)$$   The derivative can be either positive or negative, so $$f\left( x \right)$$ is not strictly monotonic on $$\left( {-\infty ,\,\infty } \right)$$, and therefore does not have an inverse. Strictly monotonic on entire domain?   $$\displaystyle f\left( x \right)=\ln \left( x \right)$$ $$f\left( x \right)=\ln \left( x \right),\,\,\,\,x>0$$ $$\displaystyle {f}’\left( x \right)=\frac{1}{x}>0,\,\,\,\,\text{for }x>0$$   The derivative is positive for all $$x$$, so $$f\left( x \right)$$ is increasing on $$\left( {0,\,\infty } \right)$$. Thus $$f\left( x \right)$$ is strictly monotonic and has an inverse.

# Finding the Derivative of an Inverse Function

The derivative of an inverse function can be found the following way; note that $$f\left( {g\left( x \right)} \right)$$ means a composite function, which means that we take the inside function, $$g\left( x \right)$$, and put that in everywhere there’s an “$$x$$” in the outside function, .$$f\left( x \right)$$

Derivative of an Inverse Function

Let $$f\left( x \right)$$ be a function that is differentiable on a certain interval. If $$f\left( x \right)$$ has an inverse function $$g\left( x \right)$$, and $$g\left( x \right)$$ is differentiable for any value of $$x$$ such that $${f}’\left( {g\left( x \right)} \right)\ne 0$$, then:

$$\displaystyle {g}’\left( x \right)=\frac{1}{{{f}’\left( {g\left( x \right)} \right)}}$$

What this says is if we have a function and want to find the derivative of the inverse of the function at a certain point “$$x$$”, we just find the “$$y$$” for the particular “$$x$$” in the original function, and use this value as the “$$x$$” in the derivative of this function. Then take the reciprocal of this number; this gives the derivative of the inverse of the original function at this point.

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Another way to explain this is “the derivative of $$f\left( x \right)$$  at a point $$(a, b)$$ is the reciprocal of the derivative of $${{f}^{{-1}}}\left( x \right)$$ at point $$(b, a)$$”.  (We’ll do problems below).

Here are some Derivative of the Inverse problems. Some teachers may have you solve these using implicit differentiation, so I’m including that method, too.

Remember that we must first check that the function is monotonic, to make sure the function is one-to-one, or has an inverse.

 Derivative of the Inverse Solution Find $${{\left( {{{f}^{{-1}}}} \right)}^{\prime }}\left( 4 \right)$$ for:   $$f\left( x \right)={{x}^{3}}-{{x}^{2}}+x+3$$     $$f$$ is monotonic (increasing) on $$\left( {-\infty ,\infty } \right)$$, so it has an inverse. Method 1: $$f\left( x \right)={{x}^{3}}-{{x}^{2}}+x+3,\,\,\,\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}-2x+1$$   Find $${{f}^{{-1}}}\left( 4 \right)$$ (find $$x$$ when $$y=4$$): $${{x}^{3}}-{{x}^{2}}+x+3=4;\,\,\,\,{{x}^{3}}-{{x}^{2}}+x-1=0$$ Let’s try 1 as a root; use synthetic division to verify: ; $${{f}^{{-1}}}\left( 4 \right)=1$$. Now, since $${{f}^{{-1}}}\left( 4 \right)=1$$, use the value 1 in the derivative function ($$3{{x}^{2}}-2x+1$$), and then take the reciprocal: $$\displaystyle {{\left( {{{f}^{{-1}}}} \right)}^{\prime }}\left( 4 \right)=\frac{1}{{{f}’\left( {{{f}^{{-1}}}\left( 4 \right)} \right)}}=\frac{1}{{{f}’\left( 1 \right)}}=\frac{1}{{3{{{\left( 1 \right)}}^{2}}-2\left( 1 \right)+1}}=\frac{1}{2}$$   Method 2 (Implicit Integration): To get $${{f}^{{-1}}}\left( x \right)$$, switch $$x$$ and $$y$$ to get the new $$y={{f}^{{-1}}}\left( x \right)$$: $$x={{y}^{3}}-{{y}^{2}}+y+3\,\,\,\to \,\,4={{y}^{3}}-{{y}^{2}}+y+3\,\,\to \,\,y=1$$   (from above) Then use implicit differentiation and get $$\displaystyle \frac{{d\left( {{{f}^{{-1}}}\left( x \right)} \right)}}{{dx}}$$: $$\displaystyle \frac{{dx}}{{dx}}=3{{y}^{2}}\frac{{dy}}{{dx}}-2y\frac{{dy}}{{dx}}+1\frac{{dy}}{{dx}}$$ $$\displaystyle \frac{{dy}}{{dx}}\left( {3{{y}^{2}}-2y+1} \right)=1$$ $$\displaystyle \frac{{dy}}{{dx}}=\frac{1}{{3{{y}^{2}}-2y+1}}=\frac{1}{{3{{{\left( 1 \right)}}^{2}}-2\left( 1 \right)+1}}=\frac{1}{2}$$ Find $${{\left( {{{f}^{{-1}}}} \right)}^{\prime }}\left( 1 \right)$$ for:   $$f\left( x \right)=2\sin x,$$ $$\displaystyle -\frac{\pi }{2}\le x\le \frac{\pi }{2}$$     $$f$$ is monotonic (increasing) on $$\displaystyle \left[ {-\frac{\pi }{2},\,\,\frac{\pi }{2}} \right]$$, so it has an inverse in that interval. Method 1: $$\displaystyle f\left( x \right)=2\sin \left( x \right),\,\,\,\,\,\,\,{f}’\left( x \right)=2\cos \left( x \right)\,\,\,\,\,\,-\frac{\pi }{2}\le x\le \frac{\pi }{2}$$:   Quadrants I and IV   Find $${{f}^{{-1}}}\left( 1 \right)$$ (find $$x$$ when $$y=1$$): $$\displaystyle 2\sin x=1;\,\,\,\,\sin x=\frac{1}{2};\,\,\,\,x={{\sin }^{{-1}}}\left( {\frac{1}{2}} \right);\,\,\,\,\,x=\frac{\pi }{6};\,\,\,\,\,\,f\left( {\frac{\pi }{6}} \right)=1$$ Now, since $$\displaystyle f\left( {\frac{\pi }{6}} \right)=1$$, use the value $$\displaystyle {\frac{\pi }{6}}$$ in the derivative function ($$\displaystyle 2\cos \left( x \right)$$), and then take the reciprocal: \displaystyle \begin{align}{{\left( {{{f}^{{-1}}}} \right)}^{\prime }}\left( 1 \right)&=\frac{1}{{{f}’\left( {{{f}^{{-1}}}\left( 1 \right)} \right)}}=\frac{1}{{{f}’\left( {\frac{\pi }{6}} \right)}}=\frac{1}{{2\cos \left( {\frac{\pi }{6}} \right)}}\\&=\frac{1}{{2\cdot \frac{{\sqrt{3}}}{2}}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}\end{align}   Method 2 (Implicit Integration): To get $${{f}^{{-1}}}\left( x \right)$$, switch $$x$$ and $$y$$ to get the new $$y={{f}^{{-1}}}\left( x \right)$$: $$\displaystyle x=2\sin y\,\,\,\to \,\,1=2\sin y\,\,\to \,\,\sin y=\frac{1}{2};\,\,\,\,y=\frac{\pi }{6}$$ for $$\displaystyle -\frac{\pi }{2}\le y\le \frac{\pi }{2}$$ Then use implicit differentiation and get $$\displaystyle \frac{{d\left( {{{f}^{{-1}}}\left( x \right)} \right)}}{{dx}}$$: $$\displaystyle \frac{{dx}}{{dx}}=2\cos y\cdot \frac{{dy}}{{dx}}\,$$ $$\displaystyle 2\cos y\cdot \frac{{dy}}{{dx}}=1$$ $$\displaystyle \frac{{dy}}{{dx}}=\frac{1}{{2\cos y}}=\frac{1}{{2\cos \left( {\frac{\pi }{6}} \right)}}=\frac{1}{{2\left( {\frac{{\sqrt{3}}}{2}} \right)}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$$

On to Antiderivatives and Indefinite Integration, including Trig Integration – you are ready!

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