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## Section2.6Derivatives of Inverse Functions

###### Motivating Questions
• What is the derivative of the natural logarithm function?

• What are the derivatives of the inverse trigonometric functions $\arcsin(x)$ and $\arctan(x)\text{?}$

• If $g$ is the inverse of a differentiable function $f\text{,}$ how is $g'$ computed in terms of $f\text{,}$ $f'\text{,}$ and $g\text{?}$

Much of mathematics centers on the notion of function. Indeed, throughout our study of calculus we are investigating the behavior of functions, with a particular emphasis on how fast the output of the function changes in response to changes in the input. Because each function represents a process, a natural question to ask is whether or not such a process can be reversed. That is, if we know the output that results from the function, can we (uniquely) determine the input that led to it? And if we know how fast a particular process is changing, can we determine how fast the inverse process is changing?

One of the most important functions in all of mathematics is the natural exponential function $f(x) = e^x\text{.}$ Its inverse, the natural logarithm $g(x) = \ln(x)\text{,}$ is similarly important. One of our goals in this section is to learn how to differentiate the logarithm function. First, we review some of the basic concepts surrounding functions and their inverses in Example2.69.

###### Example2.69

The equation $y = \frac{5}{9}(x-32)$ relates a temperature given in $x$ degrees Fahrenheit to the corresponding temperature $y$ measured in degrees Celsius.

1. Write $x$ (Fahrenheit temperature) in terms of $y$ (Celsius temperature) by solving the equation $y = \frac{5}{9}(x-32)$ for $x\text{.}$

2. Let $C(x) = \frac{5}{9}(x-32)$ be the function that takes a Fahrenheit temperature as input and produces the equivalent Celsius temperature as output. In addition, let $F(y)$ be the function that converts a temperature given in $y$ degrees Celsius to the equivalent temperature $F(y)$ measured in degrees Fahrenheit. Use your work in (a) to write a formula for $F(y)\text{.}$

3. Next consider the composite function defined by $p(x) = F(C(x))\text{.}$ Use the formulas for $F$ and $C$ to determine an expression for $p(x)$ and simplify this expression as much as possible. What do you observe?

4. Now, let $r(y) = C(F(y))\text{.}$ Use the formulas for $F$ and $C$ to determine an expression for $r(y)$ and simplify this expression as much as possible. What do you observe?

5. What is the value of $C'(x)\text{?}$ of $F'(y)\text{?}$ How do these values appear to be related?

Hint
1. Start by multiplying both sides of the equation by $\frac{9}{5}\text{.}$

2. How should $F(y)$ and $x$ be related?

3. How do the units of $x$ compare to the units of $p(x)\text{?}$

4. How do the units of $y$ compare to the units of $r(y)\text{?}$

5. The units on $C'(x)$ are degrees Celsius per degree Fahrenheit; the units on $F'(y)$ are degrees Fahrenheit per degree Celsius.

1. $x=\frac95y+32\text{.}$

2. $F(y)=\frac95y+32\text{.}$

3. $p(x)=x\text{.}$

4. $r(y)=y\text{.}$

5. $C'(x)=\frac59$ is the reciprocal of $F'(y)=\frac95\text{.}$

Solution
1. To solve the equation $y=\frac59(x-32)$ for $x\text{,}$ we first multiply both sides by $\frac95$ and then add $32\text{:}$

\begin{align*} y=\mathstrut \amp \frac59(x-32)\\ \frac95y=\mathstrut \amp x-32 \\ \frac95y+32=\mathstrut \amp x\text{.} \end{align*}
2. From the initial equation $y=\frac59(x-32)$ that relates a temperature of $x$ degrees Fahrenheit to the equivalent temperature $y$ degrees Celsius, we have said that $C(x)=y\text{.}$ Similarly, our work in part (a) rewrote that equation into the form $x=\frac95y+32$ demonstrating the same relation between a temperature of $x$ degrees Fahrenheit and $y$ degrees Celsius. It should be the case, then, that $F(y)=x\text{.}$ Thus a formula for $F$ is

\begin{equation*} F(y)=\frac95y+32\text{.} \end{equation*}
3. In part (b), we said that $C(x)=y$ and $F(y)=x\text{.}$ Based on this, it should be the case that

\begin{equation*} p(x)=F(C(x))=F(y)=x\text{,} \end{equation*}

but we'll take a closer look to double check. Observe:

\begin{align*} p(x)=\mathstrut \amp F(C(x))\\ =\mathstrut \amp \frac95C(x)+32\\ =\mathstrut \amp \frac95\left(\frac59(x-32)\right)+32\\ =\mathstrut \amp (x-32)+32\\ =\mathstrut \amp x\text{.} \end{align*}

To make sense of this, we notice that the function $p$ takes a temperature $x\text{,}$ in degrees Fahrenheit, pushes it through the function $C$ to convert it to $C(x)$ degrees Celsius, and then pushes the result through the function $F$ to convert it back into degrees Fahrenheit. Since $F$ and $C$ are each just converting the units of a given temperature, and the units of $F(C(x))$ are the same as the units on $x\text{,}$ it is perfectly reasonable to end up with the equation $x=F(C(x))\text{.}$

4. The situation with $r(y)=C(F(y))$ is analogous to that described in part (c). In part (b), we said that $C(x)=y$ and $F(y)=x\text{.}$ Based on this, it should be the case that

\begin{equation*} r(y)=C(F(y))=C(x)=y\text{,} \end{equation*}

but we'll take a closer look to double check. Observe:

\begin{align*} r(y)=\mathstrut \amp C(F(y))\\ =\mathstrut \amp \frac59(F(y)-32)\\ =\mathstrut \amp \frac59\left(\left(\frac95y+32\right)-32\right)\\ =\mathstrut \amp \frac59\left(\frac95y\right)\\ =\mathstrut \amp y\text{.} \end{align*}

To make sense of this, we notice that the function $r$ takes a temperature $y\text{,}$ in degrees Celsius, pushes it through the function $F$ to convert it to $F(y)$ degrees Fahrenheit, and then pushes the result through the function $C$ to convert it back into degrees Celsius. Since $F$ and $C$ are each just converting the units of a given temperature, and the units of $C(F(y))$ are the same as the units on $y\text{,}$ it is perfectly reasonable to end up with the equation $y=C(F(y))\text{.}$

From this and part (c), we deduce that $C$ and $F$ are a pair of inverse functions.

5. Since $C(x)=\frac59(x-32)\text{,}$ we know that $C'(x)=\frac59$ degrees Celsius per degree Fahrenheit. Likewise, since $F(y)=\frac95y+32\text{,}$ we know that $F'(y)=\frac95$ degrees Fahrenheit per degree Celsius. We note that these values are reciprocals of each other, as might be expected from the units on each.

### SubsectionBasic Facts about Inverse Functions

###### Inverse Functions

A function $f : A \to B$ is a rule that associates each element in the set $A$ to one and only one element in the set $B\text{.}$ We call $A$ the domain of $f$ and $B$ the codomain of $f\text{.}$ If there exists a function $g : B \to A$ such that $g(f(a)) = a$ for every possible choice of $a$ in the set $A$ and $f(g(b)) = b$ for every $b$ in the set $B\text{,}$ then we say that $g$ is the inverse of $f\text{.}$

We often use the notation $f^{-1}$ (read $"f$-inverse") to denote the inverse of $f\text{.}$ The inverse function undoes the work of $f\text{.}$ Indeed, if $y = f(x)\text{,}$ then

\begin{equation*} f^{-1}(y) = f^{-1}(f(x)) = x\text{.} \end{equation*}

Thus, the equations $y = f(x)$ and $x = f^{-1}(y)$ say the same thing. The only difference between the two equations is one is solved for $x\text{,}$ while the other is solved for $y\text{.}$

Here we briefly remind ourselves of some key facts about inverse functions.

###### Note2.70

For a function $f : A \to B\text{,}$

• $f$ has an inverse if and only if $f$ is one-to-one10A function $f$ is one-to-one provided that no two distinct inputs lead to the same output. and onto;11A function $f$ is onto provided that every possible element of the codomain can be realized as an output of the function for some choice of input from the domain.

• Provided $f^{-1}$ exists, the domain of $f^{-1}$ is the codomain of $f\text{,}$ and the codomain of $f^{-1}$ is the domain of $f\text{;}$

• $f^{-1}(f(x)) = x$ for every $x$ in the domain of $f$ and $f(f^{-1}(y)) = y$ for every $y$ in the codomain of $f\text{;}$

• $y = f(x)$ if and only if $x = f^{-1}(y)\text{.}$

The last fact reveals a special relationship between the graphs of $f$ and $f^{-1}\text{.}$ If a point $(x,y)$ lies on the graph of $y = f(x)\text{,}$ then it is also true that $x = f^{-1}(y)\text{,}$ which means that the point $(y,x)$ lies on the graph of $f^{-1}\text{.}$ This shows us that the graphs of $f$ and $f^{-1}$ are the reflections of each other across the line $y = x\text{,}$ because this reflection is precisely the geometric action that swaps the coordinates in an ordered pair. In Figure2.71 below, we see this illustrated by the graphs of $y=f(x)$ and $y=f^{-1}(x)\text{,}$ where $f(x) = 2^x\text{.}$ The marked points $\big(-1,\frac{1}{2}\big)$ and $\big(\frac{1}{2},-1\big)$ highlight the reflection of the curves across the line $y = x\text{.}$ Figure2.71The graph of a function $y = f(x)$ along with its inverse $y = f^{-1}(x)\text{.}$

To close our review of important facts about inverses, we recall that the natural exponential function $f(x) = e^x$ has an inverse function, namely the natural logarithm $f^{-1}(y) = \ln(y)\text{.}$ Thus, writing $y = e^x$ is interchangeable with $x = \ln(y)\text{,}$ while also $\ln(e^x) = x$ for every real number $x\text{,}$ and $e^{\ln(y)} = y$ for every positive real number $y\text{.}$

### SubsectionThe Derivative of the Natural Logarithm Function

In what follows, we find a formula for the derivative of $g(x) = \ln(x)\text{.}$ To do so, we take advantage of the fact that we know the derivative of the natural exponential function, the inverse of $g\text{.}$ In particular, we know that writing $g(x) = \ln(x)$ is equivalent to writing $e^{g(x)} = x\text{.}$ Now, we differentiate both sides of this equation and observe that

\begin{equation*} \frac{d}{dx}\left[e^{g(x)}\right] = \frac{d}{dx}[x]\text{.} \end{equation*}

The righthand side is simply $1\text{;}$ by applying the chain rule to the left side, we find that

\begin{equation*} e^{g(x)} g'(x) = 1\text{.} \end{equation*}

Next we solve for $g'(x)$ and get

\begin{equation*} g'(x) = \frac{1}{e^{g(x)}}\text{.} \end{equation*}

Finally, we recall that $g(x) = \ln(x)\text{,}$ so $e^{g(x)} = e^{\ln(x)} = x\text{.}$ Thus

\begin{equation*} g'(x) = \frac{1}{x}\text{.} \end{equation*}

###### Derivative of the Natural Logarithm

For all positive real numbers $x\text{,}$ $\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$

This rule for the natural logarithm function now joins our list of basic derivative rules. Note that this rule applies only to positive values of $x\text{,}$ as these are the only values for which $\ln(x)$ is defined.

Also notice that for the first time in our work, differentiating a basic function of a particular type has led to a function of a very different nature: the derivative of the natural logarithm is not another logarithm, nor even an exponential function, but rather a rational one.

Derivatives of logarithms may now be computed in concert with all of the rules known to date. For instance, if $f(t) = \ln(t^2 + 1)\text{,}$ then by the chain rule, $f'(t) = \frac{1}{t^2 + 1} \cdot 2t\text{.}$

There are interesting connections between the graphs of $y=f(x)$ and $y=f^{-1}(x)\text{,}$ with $f(x) = e^x$ and $f^{-1}(x) = \ln(x)\text{.}$

In Figure2.72 on the right, we are reminded that since the natural exponential function has the property that its derivative is itself, the slope of the tangent to $y = e^x$ is equal to the height of the curve at that point. For instance, at the point $A = (\ln(0.5), 0.5)\text{,}$ the slope of the tangent line is $m_A = 0.5\text{,}$ and at $B = (\ln(5), 5)\text{,}$ the tangent line's slope is $m_B = 5\text{.}$ At the corresponding points $A'$ and $B'$ on the graph of the natural logarithm function (which come from reflecting $A$ and $B$ across the line $y = x$), we know that the slope of the tangent line is the reciprocal of the $x$-coordinate of the point (since $\frac{d}{dx}[\ln(x)] = \frac{1}{x}$). Thus, at $A' = (0.5, \ln(0.5))\text{,}$ we have $m_{A'} = \frac{1}{0.5} = 2\text{,}$ and at $B' = (5, \ln(5))\text{,}$ $m_{B'} = \frac{1}{5}\text{.}$

In particular, we observe that $m_{A'} = \frac{1}{m_A}$ and $m_{B'} = \frac{1}{m_B}\text{.}$ This is not a coincidence, but in fact holds for any curve $y = f(x)$ and its inverse, provided the inverse exists. This is due to the reflection of the graphs across the line $y = x\text{:}$ the reflection changes the roles of $x$ and $y\text{,}$ thus reversing the rise and run, so the slope of the inverse function at the reflected point is the reciprocal of the slope of the original function.

###### Example2.73

For each function given below, find its derivative.

1. $h(x) = x^2\ln(x)$

2. $p(t) = \frac{\ln(t)}{e^t + 1}$

3. $s(y) = \ln(\cos(y) + 2)$

4. $z(x) = \tan(\ln(x))$

5. $m(z) = \ln(\ln(z))$

Hint
1. Is $h$ a product, quotient, or composition of basic functions?

2. Is $p$ a product, quotient, or composition of basic functions?

3. Is $s$ a product, quotient, or composition of basic functions?

4. Is $z$ a product, quotient, or composition of basic functions?

5. Is $m$ a product, quotient, or composition of basic functions?

1. $h'(x) = 2x\ln(x)+x\text{.}$

2. $\displaystyle p'(t) = \frac{ \frac{1}{t}(e^t + 1) - \ln(t) e^t}{(e^t + 1)^2}\text{.}$

3. $s'(y) = \frac{1}{\cos(y) + 2)} \cdot (-\sin(y))\text{.}$

4. $z'(x) = \frac{\sec^2(\ln(x))}{x}\text{.}$

5. $m'(z) = \frac{1}{z\ln(z)}\text{.}$

Solution
1. By the product rule,

\begin{equation*} h'(x) = 2x\ln(x)+x^2\cdot\frac{1}{x} = 2x\ln(x)+x\text{.} \end{equation*}
2. By the quotient rule,

\begin{equation*} p'(t) = \frac{\frac{1}{t}(e^t + 1) - \ln(t) e^t}{(e^t + 1)^2}\text{.} \end{equation*}
3. The chain rule tells us that

\begin{equation*} s'(y) = \frac{1}{\cos(y) + 2)} \cdot (-\sin(y))\text{.} \end{equation*}
4. Again using the chain rule,

\begin{equation*} z'(x) = \sec^2(\ln(x)) \cdot \frac{1}{x}\text{.} \end{equation*}
5. Noting that $m$ is composite with the natural logarithm function serving as both the inner and outer function, we find that

\begin{equation*} m'(z) = \frac{1}{\ln(z)} \cdot \frac{d}{dz}\left[\ln(z)\right]=\frac{1}{z\ln(z)}\text{.} \end{equation*}

### SubsectionInverse Trigonometric Functions and their Derivatives

Trigonometric functions are periodic, so they fail to be one-to-one, and thus do not have inverse functions. However, we can restrict the domain of each trigonometric function so that it is one-to-one on that domain.

For instance, consider the sine function on the domain $[-\frac{\pi}{2}, \frac{\pi}{2}]\text{.}$ Because no output of the sine function is repeated on this interval, the function is one-to-one and thus has an inverse. Thus, the function $f(x) = \sin(x)$ with domain $\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]$ and codomain $[-1,1]$ has an inverse function $f^{-1}$ such that

\begin{equation*} f^{-1} : [-1,1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\text{.} \end{equation*}

We call $f^{-1}$ the arcsine (or inverse sine) function and write $f^{-1}(y) = \arcsin(y)\text{.}$ It is especially important to remember that

\begin{equation*} y = \sin(x) \ \ \text{and} \ \ x = \arcsin(y) \end{equation*}

say the same thing. The arcsine of $y$ means the angle whose sine is $y\text{.}$ For example, $\arcsin\big(\frac{1}{2}\big) = \frac{\pi}{6}$ means that $\frac{\pi}{6}$ is the angle whose sine is $\frac{1}{2}\text{,}$ which is equivalent to writing $\sin\big(\frac{\pi}{6}\big) = \frac{1}{2}\text{.}$ Next, we determine the derivative of the arcsine function. Letting $h(x) = \arcsin(x)\text{,}$ our goal is to find $h'(x)\text{.}$ Since $h(x)$ is the angle whose sine is $x\text{,}$ it is equivalent to write

\begin{equation*} \sin(h(x)) = x\text{.} \end{equation*}

Differentiating both sides of the previous equation, we have

\begin{equation*} \frac{d}{dx}[\sin(h(x))] = \frac{d}{dx}[x]\text{.} \end{equation*}

The righthand side is simply $1\text{,}$ and by applying the chain rule to the left side, we end up with

\begin{equation*} \cos(h(x)) h'(x) = 1\text{.} \end{equation*}

Solving for $h'(x)\text{,}$ it follows that

\begin{equation*} h'(x) = \frac{1}{\cos(h(x))}\text{.} \end{equation*}

Finally, we recall that $h(x) = \arcsin(x)\text{,}$ so the denominator of $h'(x)$ is the function $\cos(\arcsin(x))\text{,}$ or in other words, the cosine of the angle whose sine is $x\text{:}$

\begin{equation*} h'(x)=\frac1{\cos(\arcsin(x))}\text{.} \end{equation*}

A bit of right triangle trigonometry allows us to simplify this expression considerably.

Let's say that $\theta = \arcsin(x)\text{,}$ so that $\theta$ is the angle whose sine is $x\text{.}$ We can picture $\theta$ as an angle in a right triangle with hypotenuse $1$ and a vertical leg of length $x\text{,}$ as shown on the right in Figure2.75. By the Pythagorean Theorem, the length of the horizontal leg must be $\sqrt{1-x^2}\text{.}$ Now, because $\theta = \arcsin(x)\text{,}$ the expression for $\cos(\arcsin(x))$ is equivalent to $\cos(\theta)\text{.}$ From the figure, $\cos(\arcsin(x)) = \cos(\theta) = \sqrt{1-x^2}\text{.}$

Substituting this expression into our formula for $h'(x)\text{,}$ we have now shown that

\begin{equation*} h'(x) = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}

###### Derivative of the Inverse Sine

For all real numbers $x$ such that $-1 \lt x \lt 1\text{,}$

\begin{equation*} \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}

###### Example2.76

The following prompts in this example will lead you to develop the derivative of the inverse tangent function.

1. Let $r(x) = \arctan(x)\text{.}$ Use the relationship between the arctangent and tangent functions to rewrite this equation using only the tangent function.

2. Differentiate both sides of the equation you found in (a). Solve the resulting equation for $r'(x)$ and write $r'(x)$ as simply as possible in terms of a trigonometric function evaluated at $r(x)\text{.}$

3. Recall that $r(x) = \arctan(x)\text{.}$ Update your expression for $r'(x)$ so that it only involves trigonometric functions and the independent variable $x\text{.}$

4. Introduce a right triangle with angle $\theta$ so that $\theta = \arctan(x)\text{.}$ What are the three sides of the triangle?

5. In terms of only $x$ and $1\text{,}$ what is the value of $\cos(\arctan(x))\text{?}$

6. Use the results of your work above to find an expression involving only $1$ and $x$ for $r'(x)\text{.}$

Hint
1. Recall that for any function and its inverse, writing $y = f^{-1}(x)$ is equivalent to writing $f(y) = x\text{.}$

2. Apply the chain rule to differentiate $\tan(r(x))\text{.}$

3. This question is only asking you to substitute the expression for $r(x)$ into what you found in (b).

4. Let the vertical leg of the triangle be $x\text{.}$ What must the horizontal leg be?

5. Recall that the cosine of an angle is the length of the adjacent leg over the length of the hypotenuse.

6. Note that $\cos^2(\alpha) = (\cos(\alpha))^2\text{.}$

1. $\tan(r(x)) = x\text{.}$

2. $r'(x) = \cos^2(r(x))\text{.}$

3. $r'(x) = \cos^2(\arctan(x))\text{.}$

4. With $\theta = \arctan(x)\text{,}$ 5. $\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.}$

6. $r'(x) = \frac{1}{1+x^2}\text{.}$

Solution
1. Since $r(x) = \arctan(x)\text{,}$ it is equivalent to write $\tan(r(x)) = x\text{.}$

2. Differentiating, we have $\frac{d}{dx}[\tan(r(x))] = \frac{d}{dx}[x]\text{,}$ so

\begin{equation*} \sec^2(r(x)) r'(x) = 1\text{,} \end{equation*}

and thus

\begin{equation*} r'(x) = \frac{1}{\sec^2(r(x))} = \cos^2(r(x))\text{,} \end{equation*}

since $\frac{1}{\sec(\alpha)} = \cos(\alpha)\text{.}$

3. Since $r(x) = \arctan(x)\text{,}$ we now have that $r'(x) = \cos^2(\arctan(x))\text{.}$

4. Letting $\theta = \arctan(x)\text{,}$ it follows that we can view $\theta$ as an angle in a right triangle with legs $1$ and $x$ (so that $\tan(\theta) = \frac{x}{1}\text{,}$ and thus by the Pythagorean Theorem, the triangle's hypotenuse is $\sqrt{1+x^2}\text{,}$ as shown below. 5. To evaluate $\cos(\arctan(x))\text{,}$ we use the right triangle developed in (d) and observe that

\begin{equation*} \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.} \end{equation*}
6. Finally, we recall that we know $r'(x) = \cos^2(\arctan(x)) = \left( \cos(\arctan(x)) \right)^2\text{.}$ Having established that $\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{,}$ we now have that

\begin{equation*} r'(x) = \frac{1}{1+x^2}\text{.} \end{equation*}

###### Derivative of the Inverse Tangent

For all real numbers $x\text{,}$

\begin{equation*} \frac{d}{dx}[\arctan(x)]=\frac1{1+x^2}\text{.} \end{equation*}

While derivatives for other inverse trigonometric functions can be established similarly, for now we limit ourselves to the arcsine and arctangent functions.

###### Example2.77

Determine the derivative of each of the following functions.

1. $\displaystyle f(x) = x^3 \arctan(x) + e^x \ln(x)$

2. $\displaystyle p(t) = 2^{t\arcsin(t)}$

3. $\displaystyle h(z) = (\arcsin(5z) + \arctan(4-z))^{27}$

4. $\displaystyle s(y) = \cot(\arctan(y))$

5. $\displaystyle m(v) = \ln(\sin^2(v)+1)$

6. $\displaystyle g(w) = \arctan\left( \frac{\ln(w)}{1+w^2} \right)$

Hint
1. Use the sum rule, followed by the product rule on each term in the sum.

2. Use the chain rule first. What rule is needed to differentiate $t\arcsin(t)\text{?}$

3. Note that the chain rule is needed to differentiate $\arcsin(5z)\text{.}$

4. You can use right triangle trigonometry to simplify the function first.

5. Use the chain rule twice.

6. Think about the overall structure of this function.

1. $f'(x) = \left[3x^2\arctan(x)+x^3 \cdot \frac{1}{1+x^2}\right] + \left[e^x\ln(x)+e^x \cdot \frac{1}{x}\right]\text{.}$

2. $p'(t) = 2^{t\arcsin(t)} \ln(2) \left[\arcsin(t)+t \cdot \frac{1}{\sqrt{1-t^2}}\right]\text{.}$

3. $h'(z) = 27\left(\arcsin(5z) + \arctan(4-z)\right)^{26} \left[\frac{1}{\sqrt{1-(5z)^2}} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right]\text{.}$

4. $s'(y) = -\frac{1}{y^2}\text{.}$

5. $m'(v) = \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right]\text{.}$

6. $g'(w) = \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{\frac{1}{w}(1+w^2) - \ln(w) 2w}{(1+w^2)^2} \right] \text{.}$

Solution
1. By the sum rule followed by two applications of the product rule,

\begin{align*} f'(x) =\mathstrut \amp \frac{d}{dx}[x^3 \arctan(x)] + \frac{d}{dx}[e^x \ln(x)]\\ =\mathstrut \amp \left[3x^2\arctan(x)+x^3 \cdot \frac{1}{1+x^2}\right] + \left[e^x\ln(x)+e^x \cdot \frac{1}{x}\right]\text{.} \end{align*}
2. Using the chain rule followed by the product rule,

\begin{align*} p'(t) =\mathstrut \amp 2^{t\arcsin(t)} \ln(2) \frac{d}{dt}[t\arcsin(t)]\\ =\mathstrut \amp 2^{t\arcsin(t)} \ln(2) \left[\arcsin(t)+t \cdot \frac{1}{\sqrt{1-t^2}}\right]\text{.} \end{align*}
3. Applying the chain rule twice,

\begin{align*} h'(z) =\mathstrut \amp 27\big(\arcsin(5z) + \arctan(4-z)\big)^{26} \frac{d}{dz}\left[\arcsin(5z) + \arctan(4-z)\right]\\ =\mathstrut \amp 27\left(\arcsin(5z) + \arctan(4-z)\right)^{26} \left[\frac{1}{\sqrt{1-(5z)^2}} \cdot \frac{d}{dz}[5z] + \frac{1}{1+(4-z)^2} \cdot \frac{d}{dz}[4-z] \right]\\ =\mathstrut \amp 27\left(\arcsin(5z) + \arctan(4-z)\right)^{26} \left[\frac{1}{\sqrt{1-(5z)^2}} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right]\text{.} \end{align*}
4. Using right triangle trigonometry, it is straightforward to show that $\cot(\arctan(y)) = \frac{1}{y}\text{.}$ As such, $s'(y) = -\frac{1}{y^2}\text{.}$

5. By two applications of the chain rule (noting particularly that $\sin^2(v) = (\sin(v))^2\text{,}$ we have

\begin{align*} m'(v) =\mathstrut \amp \frac{1}{\sin^2(v)+1} \cdot \frac{d}{dv} \left[ \sin^2(v) + 1 \right]\\ =\mathstrut \amp \frac{1}{\sin^2(v)+1}\cdot \left[2\sin(v)\frac{d}{dv}[\sin(v)]\right]\\ =\mathstrut \amp \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right]\text{.} \end{align*}
6. By the chain rule followed by the quotient rule,

\begin{align*} g'(w) =\mathstrut \amp \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \frac{d}{dw} \left[ \frac{\ln(w)}{1+w^2} \right]\\ =\mathstrut \amp \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{\frac{1}{w}(1+w^2) - \ln(w) 2w}{(1+w^2)^2} \right]\text{.} \end{align*}

### SubsectionThe Link Between the Derivative of a Function and the Derivative of its Inverse

Earlier, we saw in Figure2.72 an interesting relationship between the slopes of tangent lines to the natural exponential and natural logarithm functions at points reflected across the line $y = x\text{.}$ In particular, we observed that at the point $(\ln(2), 2)$ on the graph of $f(x) = e^x\text{,}$ the slope of the tangent line is $f'(\ln(2)) = 2\text{,}$ while at the corresponding point $(2, \ln(2))$ on the graph of $f^{-1}(x) = \ln(x)\text{,}$ the slope of the tangent line is $(f^{-1})'(2) = \frac{1}{2}\text{,}$ which is the reciprocal of $f'(\ln(2))\text{.}$

That the two corresponding tangent lines have reciprocal slopes is not a coincidence. If $f$ and $g$ are differentiable inverse functions, then $y = f(x)$ if and only if $x = g(y)\text{,}$ and thus $f(g(x)) = x$ for every $x$ in the domain of $f^{-1}\text{.}$ Differentiating both sides of this equation, we have

\begin{equation*} \frac{d}{dx} [f(g(x))] = \frac{d}{dx} [x]\text{,} \end{equation*}

and by the chain rule,

\begin{equation*} f'(g(x)) g'(x) = 1\text{.} \end{equation*}

Solving for $g'(x)\text{,}$ we have $g'(x) = \frac{1}{f'(g(x))}\text{.}$ Here we see that the slope of the tangent line to the inverse function $g$ at the point $(x,g(x))$ is precisely the reciprocal of the slope of the tangent line to the original function $f$ at the point $(g(x),f(g(x))) = (g(x),x)\text{.}$ Figure2.78A graph of a function $y = f(x)$ along with its inverse $y = g(x) = f^{-1}(x)\text{.}$ Observe that the slopes of the two tangent lines are reciprocals of one another.

To see this more clearly, consider the graph of the function $y = f(x)$ shown above in Figure2.78 along with its inverse $y = g(x)\text{.}$ Given a point $(a,b)$ that lies on the graph of $f\text{,}$ we know that $(b,a)$ lies on the graph of $g$ because $f(a) = b$ and $g(b) = a\text{.}$ Applying the rule that $g'(x) = \frac{1}{f'(g(x))}$ to the value $x = b$ yields the equation

\begin{equation*} g'(b) = \frac{1}{f'(g(b))} = \frac{1}{f'(a)}\text{,} \end{equation*}

which is precisely what we see in the figure: the slope of the tangent line to $g$ at $(b,a)$ is the reciprocal of the slope of the tangent line to $f$ at $(a,b)\text{,}$ since these two lines are reflections of one another across the line $y = x\text{.}$ More generally, for any $x$ in the domain of $g'\text{,}$ we have $g'(x) = \frac{1}{f'(g(x))}\text{.}$

###### Derivative of an Inverse Function

Suppose that $f$ is a differentiable function with inverse $f^{-1}$ and that $f'(f^{-1}(x)) \ne 0\text{.}$ Then

\begin{equation*} \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}\text{.} \end{equation*}

The rules we derived for $\ln(x)\text{,}$ $\arcsin(x)\text{,}$ and $\arctan(x)$ are all just specific examples of this general property of the derivative of an inverse function. For example, with $g(x) = \ln(x)$ and $f(x) = e^x\text{,}$ it follows that

\begin{equation*} g'(x) = \frac{1}{f'(g(x))} = \frac{1}{e^{\ln(x)}} = \frac{1}{x}\text{.} \end{equation*}
###### Example2.79

Suppose $f$ is an invertible function described by Table2.80 below.

Then

\begin{equation*} \left. \frac{df^{-1}}{dx}\right\vert_{x=1}=\frac{1}{f'(f^{-1}(1))}. \end{equation*}

Now, $f^{-1}(1)=2$ and $f'(2)=4\text{.}$ Therefore,

\begin{equation*} \left. \frac{df^{-1}}{dx}\right\vert_{x=1}=\frac{1}{f'(f^{-1}(1))}=\frac{1}{f'(2)}=\frac{1}{4}. \end{equation*}

### SubsectionSummary

• For all positive real numbers $x\text{,}$ $\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$

• For all real numbers $x$ such that $-1 \lt x \lt 1\text{,}$ $\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}\text{.}$ In addition, for all real numbers $x\text{,}$ $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}\text{.}$

• If $g$ is the inverse of a differentiable function $f\text{,}$ then for any point $x$ in the domain of $g'\text{,}$ $g'(x) = \frac{1}{f'(g(x))}\text{.}$

### SubsectionExercises

Determine the derivative of each of the following functions. Use proper notation and clearly identify the derivative rules you use.

1. $f(x) = \ln(2\arctan(x) + 3\arcsin(x) + 5)$

2. $r(z) = \arctan(\ln(\arcsin(z)))$

3. $q(t) = \arctan^2(3t) \arcsin^4(7t)$

4. $g(v) = \ln\left( \frac{\arctan(v)}{\arcsin(v) + v^2} \right)$

Consider the graph of $y = f(x)$ provided in Figure2.81 and use it to answer the following questions.

1. Use the provided graph to estimate the value of $f'(1)\text{.}$

2. Sketch an approximate graph of $y = f^{-1}(x)\text{.}$ Label at least three distinct points on the graph that correspond to three points on the graph of $f\text{.}$

3. Based on your work in (a), what is the value of $(f^{-1})'(-1)\text{?}$ Why? Let $f(x) = \frac{1}{4}x^3 + 4\text{.}$

1. Sketch a graph of $y = f(x)$ and explain why $f$ is an invertible function.

2. Let $g$ be the inverse of $f$ and determine a formula for $g\text{.}$

3. Compute $f'(x)\text{,}$ $g'(x)\text{,}$ $f'(2)\text{,}$ and $g'(6)\text{.}$ What is the special relationship between $f'(2)$ and $g'(6)\text{?}$ Why?

Let $h(x) = x + \sin(x)\text{.}$

1. Sketch a graph of $y = h(x)$ and explain why $h$ must be invertible.

2. Explain why it does not appear to be algebraically possible to determine a formula for $h^{-1}\text{.}$

3. Observe that the point $(\frac{\pi}{2}, \frac{\pi}{2} + 1)$ lies on the graph of $y = h(x)\text{.}$ Determine the value of $(h^{-1})'(\frac{\pi}{2} + 1)\text{.}$