Derivatives of Inverse Functions

Reviewing Inverses and Functions	Monotonic Functions in an Interval
Finding the Derivative of an Inverse Function

Around the time you’re studying Exponential and Logarithmic Differentiation, you’ll probably learn how to get the derivative of an inverse function. The reason we might need to do this is it might be much easier to get the derivative of an inverse function than to take the inverse of a function and then take the derivative! In fact, 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:

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.

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 is found using the equation below; it looks really intimidating, but if you take it step-by-step, it’s not that bad. Note that $ f\left( {g\left( x \right)} \right)$ means a composite function (which we learned about here in the Advanced Functions: Compositions, Even and Odd, and Extrema section), 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)}}$ (also seen as $ \displaystyle {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( x \right)=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( x \right)} \right]}}$ or $ \displaystyle {{\left[ {{{f}^{{-1}}}\left( x \right)} \right]}^{\prime }}=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\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$”, first find the “$ y$” for the particular “$ x$” in the original function. Then, use this value as the “$ x$” in the derivative of this function and finally take the reciprocal of this number. This gives the derivative of the inverse of the original function at this point. 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)$”. I know this makes no sense at this point, but we’ll do problems below.

Remember that we must first check that the function is monotonic in the given interval, to make sure the function is one-to-one (has an inverse).

Here are some Derivative of the Inverse problems; notice how they may be presented in different ways. Some teachers may have you solve these using Implicit Differentiation, so I’m including that method, too. (You may also want to review how to find roots of polynomials in the Graphing and Finding Roots of Polynomial Functions section.)

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: $ \displaystyle {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( 4 \right)=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( 4 \right)} \right]}}$, $ f\left( x \right)={{x}^{3}}-{{x}^{2}}+x+3,\,\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}-2x+1$ First, 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$. Try $ 1$ as a root; use synthetic division to verify that it is in fact a root, since we get a remainder of $ 0$: ; $ {{f}^{{-1}}}\left( 4 \right)=1$. This may take some time to find a root, or you can use a graphing calculator. There should only be one root since it’s a monotonic function. 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}}=\frac{{dy}}{{dx}}={y}’$: $ \displaystyle \begin{align}x&={{y}^{3}}-{{y}^{2}}+y+3\,\\1&=3{{y}^{2}}{y}’-2y{y}’+1{y}’+0\\{y}’&\left( {3{{y}^{2}}-2y+1} \right)=1\\{y}’&=\frac{1}{{3{{y}^{2}}-2y+1}}=\frac{1}{{3{{{\left( 1 \right)}}^{2}}-2\left( 1 \right)+1}}=\frac{1}{2}\end{align}$
Let $ f$ be the function defined by $ f\left( x \right)={{x}^{3}}+x$. If $ g\left( x \right)={f}’\left( x \right)$, what is the value of $ {g}’\left( {10} \right)$? $ f$ is monotonic (increasing) on $ \left( {-\infty ,\infty } \right)$, so it has an inverse.	Method 1: $ \displaystyle {g}’\left( {10} \right)=\frac{1}{{{f}’\left[ {g\left( {10} \right)} \right]}}$, $ f\left( x \right)={{x}^{3}}+x,\,\,\,\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}+1$ First, find $ g\left( {10} \right)={{f}^{{-1}}}\left( {10} \right)$ (find what $ x$ is when $ y=10$): $ {{x}^{3}}+x=10$. Either graph, or notice that $ g\left( {10} \right)={{f}^{{-1}}}\left( {10} \right)=2$; you can verify with synthetic division. Now, since $ g\left( {10} \right)=2$, use the value $ 2$ in the derivative function ($ 3{{x}^{2}}+1$), and then take the reciprocal: $ \displaystyle {{g}^{\prime }}\left( {10} \right)=\frac{1}{{{f}’\left[ {g\left( {10} \right)} \right]}}=\frac{1}{{{f}’\left( 2 \right)}}=\frac{1}{{3{{{\left( 2 \right)}}^{2}}+1}}=\frac{1}{{13}}$ Method 2 (implicit integration): To get $ g\left( x \right)={{f}^{{-1}}}\left( x \right)$, switch $ x$ and $ y$ to get the new $ y={{f}^{{-1}}}\left( x \right)$: $ x={{y}^{3}}+y\to \,\,10={{y}^{3}}+y\,\,\to \,\,y=2$ (from above). Then, use implicit differentiation and get $ \displaystyle \frac{{d\left[ {g\left( x \right)} \right]}}{{dx}}=\frac{{d\left[ {{{f}^{{-1}}}\left( x \right)} \right]}}{{dx}}=\frac{{dy}}{{dx}}={y}’$: $ \displaystyle \begin{align}x&={{y}^{3}}+y\,\\1&=3{{y}^{2}}{y}’+1{y}’\\{y}’&\left( {3{{y}^{2}}+1} \right)=1\\{y}’&=\frac{1}{{3{{y}^{2}}+1}}=\frac{1}{{3{{{\left( 2 \right)}}^{2}}+1}}=\frac{1}{{13}}\end{align}$
Find $ {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( a \right)$ for: $ f\left( x \right)=2\sin x; \,\,a=1$ $ \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 {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( a \right)=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( a \right)} \right]}}$, $ \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 First, find $ {{f}^{{-1}}}\left( a \right),\,\,a=1$ (find what $ x$ is 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}}=\frac{{dy}}{{dx}}={y}’$: $ \displaystyle \begin{align}x&=2\sin y\\1&=2\cos y\cdot {y}’\,\\\,{y}’&=\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}\end{align}$
a) Let $ f$ and $ g$ be functions that are differentiable everywhere. If $ g$ is the inverse function of $ f$ and if $ g\left( {-3} \right)=5$ and $ \displaystyle {f}’\left( 5 \right)=-\frac{1}{4}$, then $ {g}’\left( {-3} \right)=?$ b) Let $ f$ be a differentiable function such that $ f\left( {-1} \right)=3$, $ f\left( 3 \right)=4$, and $ {f}’\left( {-1} \right)=5$, what is the value of $ {{\left[ {{{f}^{{-1}}}\left( 3 \right)} \right]}^{\prime }}$?	a) These seem really tricky, but use the equation for the derivative of an inverse: $ \displaystyle {g}’\left( x \right)=\frac{1}{{{f}’\left[ {g\left( x \right)} \right]}}:\,\,\,{g}’\left( {-3} \right)=\frac{1}{{{f}’\left[ {g\left( {-3} \right)} \right]}}=\frac{1}{{{f}’\left( 5 \right)}}=\frac{1}{{-\frac{1}{4}}}=-4$ b) Note that if $ f\left( {-1} \right)=3$, then $ {{f}^{{-1}}}\left( 3 \right)=-1$. $ \displaystyle {{\left[ {{{f}^{{-1}}}\left( x \right)} \right]}^{\prime }}=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( x \right)} \right]}}:\,\,\,{{\left[ {{{f}^{{-1}}}\left( 3 \right)} \right]}^{\prime }}=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( 3 \right)} \right]}}=\frac{1}{{{f}’\left( {-1} \right)}}=\frac{1}{5}$

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:

$ \displaystyle {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( 4 \right)=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( 4 \right)} \right]}}$, $ f\left( x \right)={{x}^{3}}-{{x}^{2}}+x+3,\,\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}-2x+1$

First, 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$. Try $ 1$ as a root; use synthetic division to verify that it is in fact a root, since we get a remainder of $ 0$: ; $ {{f}^{{-1}}}\left( 4 \right)=1$. This may take some time to find a root, or you can use a graphing calculator. There should only be one root since it’s a monotonic function.

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}}=\frac{{dy}}{{dx}}={y}’$:

$ \displaystyle \begin{align}x&={{y}^{3}}-{{y}^{2}}+y+3\,\\1&=3{{y}^{2}}{y}’-2y{y}’+1{y}’+0\\{y}’&\left( {3{{y}^{2}}-2y+1} \right)=1\\{y}’&=\frac{1}{{3{{y}^{2}}-2y+1}}=\frac{1}{{3{{{\left( 1 \right)}}^{2}}-2\left( 1 \right)+1}}=\frac{1}{2}\end{align}$

Let $ f$ be the function defined by $ f\left( x \right)={{x}^{3}}+x$. If $ g\left( x \right)={f}’\left( x \right)$, what is the value of $ {g}’\left( {10} \right)$?

$ f$ is monotonic (increasing) on $ \left( {-\infty ,\infty } \right)$, so it has an inverse.

Method 1:

$ \displaystyle {g}’\left( {10} \right)=\frac{1}{{{f}’\left[ {g\left( {10} \right)} \right]}}$, $ f\left( x \right)={{x}^{3}}+x,\,\,\,\,\,\,\,\,{f}’\left( x \right)=3{{x}^{2}}+1$

First, find $ g\left( {10} \right)={{f}^{{-1}}}\left( {10} \right)$ (find what $ x$ is when $ y=10$): $ {{x}^{3}}+x=10$. Either graph, or notice that $ g\left( {10} \right)={{f}^{{-1}}}\left( {10} \right)=2$; you can verify with synthetic division.

Now, since $ g\left( {10} \right)=2$, use the value $ 2$ in the derivative function ($ 3{{x}^{2}}+1$), and then take the reciprocal:

$ \displaystyle {{g}^{\prime }}\left( {10} \right)=\frac{1}{{{f}’\left[ {g\left( {10} \right)} \right]}}=\frac{1}{{{f}’\left( 2 \right)}}=\frac{1}{{3{{{\left( 2 \right)}}^{2}}+1}}=\frac{1}{{13}}$

Method 2 (implicit integration):

To get $ g\left( x \right)={{f}^{{-1}}}\left( x \right)$, switch $ x$ and $ y$ to get the new $ y={{f}^{{-1}}}\left( x \right)$: $ x={{y}^{3}}+y\to \,\,10={{y}^{3}}+y\,\,\to \,\,y=2$ (from above).

Then, use implicit differentiation and get $ \displaystyle \frac{{d\left[ {g\left( x \right)} \right]}}{{dx}}=\frac{{d\left[ {{{f}^{{-1}}}\left( x \right)} \right]}}{{dx}}=\frac{{dy}}{{dx}}={y}’$:

$ \displaystyle \begin{align}x&={{y}^{3}}+y\,\\1&=3{{y}^{2}}{y}’+1{y}’\\{y}’&\left( {3{{y}^{2}}+1} \right)=1\\{y}’&=\frac{1}{{3{{y}^{2}}+1}}=\frac{1}{{3{{{\left( 2 \right)}}^{2}}+1}}=\frac{1}{{13}}\end{align}$

Find $ {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( a \right)$ for:

$ f\left( x \right)=2\sin x; \,\,a=1$

$ \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 {{\left[ {{{f}^{{-1}}}} \right]}^{\prime }}\left( a \right)=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( a \right)} \right]}}$, $ \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

First, find $ {{f}^{{-1}}}\left( a \right),\,\,a=1$ (find what $ x$ is 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}}=\frac{{dy}}{{dx}}={y}’$:

$ \displaystyle \begin{align}x&=2\sin y\\1&=2\cos y\cdot {y}’\,\\\,{y}’&=\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}\end{align}$

a) Let $ f$ and $ g$ be functions that are differentiable everywhere. If $ g$ is the inverse function of $ f$ and if $ g\left( {-3} \right)=5$ and $ \displaystyle {f}’\left( 5 \right)=-\frac{1}{4}$, then $ {g}’\left( {-3} \right)=?$

b) Let $ f$ be a differentiable function such that $ f\left( {-1} \right)=3$, $ f\left( 3 \right)=4$, and $ {f}’\left( {-1} \right)=5$, what is the value of $ {{\left[ {{{f}^{{-1}}}\left( 3 \right)} \right]}^{\prime }}$?

a) These seem really tricky, but use the equation for the derivative of an inverse:

$ \displaystyle {g}’\left( x \right)=\frac{1}{{{f}’\left[ {g\left( x \right)} \right]}}:\,\,\,{g}’\left( {-3} \right)=\frac{1}{{{f}’\left[ {g\left( {-3} \right)} \right]}}=\frac{1}{{{f}’\left( 5 \right)}}=\frac{1}{{-\frac{1}{4}}}=-4$

b) Note that if $ f\left( {-1} \right)=3$, then $ {{f}^{{-1}}}\left( 3 \right)=-1$.

$ \displaystyle {{\left[ {{{f}^{{-1}}}\left( x \right)} \right]}^{\prime }}=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( x \right)} \right]}}:\,\,\,{{\left[ {{{f}^{{-1}}}\left( 3 \right)} \right]}^{\prime }}=\frac{1}{{{f}’\left[ {{{f}^{{-1}}}\left( 3 \right)} \right]}}=\frac{1}{{{f}’\left( {-1} \right)}}=\frac{1}{5}$

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