Section4.8Derivatives of Inverse Functions

Suppose we wanted to find the derivative of the inverse, but do not have an actual formula for the inverse function? Then we can use the following derivative formula for the inverse evaluated at $a\text{.}$

To see why this is true, start with the function $y=f^{-1}(x)\text{.}$ Write this as $x=f(y)$ and differentiate both sides implicitly with respect to $x$ using the Chain Rule:

\begin{equation*} 1=f'(y)\cdot \frac{dy}{dx}\text{.} \end{equation*}

Thus,

\begin{equation*} \frac{dy}{dx}=\frac{1}{f'(y)}\text{,} \end{equation*}

but $y=f^{-1}(x)\text{,}$ thus,

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

At the point $x=a$ this becomes:

\begin{equation*} \left[f^{-1}\right]'(a)=\frac{1}{f'\left[ f^{-1}(a) \right]} \end{equation*}
Example4.81. Derivatives of Inverse Functions.

Suppose $f(x)=x^5+2x^3+7x+1\text{.}$ Find $\left[f^{-1}\right]'(1)\text{.}$

Solution

First we should show that $f^{-1}$ exists (i.e. that $f$ is one-to-one). In this case the derivative $f'(x)=5x^4+6x^2+7$ is strictly greater than 0 for all $x\text{,}$ so $f$ is strictly increasing and thus one-to-one.

It's difficult to find the inverse of $f(x)$ (and then take the derivative). Thus, we use the above formula evaluated at $1\text{:}$

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

Note that to use this formula we need to know what $f^{-1}(1)$ is, and the derivative $f'(x)\text{.}$ To find $f^{-1}(1)$ we make a table of values (plugging in $x=-3,-2,-1,0,1,2,3$ into $f(x)$) and see what value of $x$ gives $1\text{.}$ We omit the table and simply observe that $f(0)=1\text{.}$ Thus,

\begin{equation*} f^{-1}(1)=0\text{.} \end{equation*}

Now we have:

\begin{equation*} \left[f^{-1}\right]'(1)=\frac{1}{f'\left( 0 \right)}\text{.} \end{equation*}

And so, $f'(0)=7\text{.}$ Therefore,

\begin{equation*} \left[f^{-1}\right]'(1)=\frac{1}{7}\text{.} \end{equation*}
Example4.82. Tangent Line of Inverse Functions.

Find the equation of the tangent line to the inverse of

\begin{equation*} f(x)=\frac{e^{-3x}}{x^{2}+1} \end{equation*}

at $(-1,0)\text{.}$

Solution

First, we differentiate

\begin{equation*} \begin{split} f'(x) \amp = \frac{(x^{2}+1)(3e^{-3x})(2x)}{(x^{2}+1)^{2}}\\ \amp = \frac{3x^{2}e^{-3x}+2xe^{-3x}+3e^{-3x}}{x^{4}+2x^{2}+1} \end{split} \end{equation*}

The derivative of $f^{-1}$ is:

\begin{equation*} \begin{split} \frac{d}{dx}f^{-1}(x) \amp = \frac{1}{\frac{3y^{2}e^{-3y}+2ye^{-3y}+3e^{-3y}}{y^{4}+2y^{2}+1}}\\[1ex] \amp = \frac{y^{4}+2y^{2}+1}{3y^{2}e^{-3y}+2ye^{-3y}+3e^{-3y}}\\[1ex] \frac{d}{dx}f^{-1}(-1) \amp = {0^{4}+0^{2}+1}{0^{2}e^{0}+2ye^{0}+3e^{0}}\\[1ex] \amp = \frac{1}{3} \end{split} \end{equation*}

An equation of the tangent line at $(-1,0)$ is then

\begin{equation*} y = \frac{1}{3}(x+1)\text{.} \end{equation*}

Subsection4.8.1Derivatives of Inverse Trigonometric Functions

We can apply the technique used to find the derivative of $f^{-1}$ above to find the derivatives of the inverse trigonometric functions.

In the following examples we will derive the formulae for the derivative of the inverse sine, inverse cosine and inverse tangent. The other three inverse trigonometric functions have been left as exercises at the end of this section.

Example4.83. Derivative of Inverse Sine.

Find the derivative of $\sin^{-1}(x)\text{.}$

Solution

As above, we write $y=\sin^{-1}(x)\text{,}$ so $x=\sin(y)$ and $-\pi/2\leq y\leq \pi/2\text{,}$ and differentiate both sides with respect to $x$ using the Chain Rule.

\begin{equation*} \begin{gathered} \ds\frac{d}{dx}x =\frac{d}{dx}\sin(y)\\ 1 =\cos(y)\frac{dy}{dx}\\ 1 =\cos\left(\sin^{-1}(x)\right)\frac{dy}{dx}\\ \frac{dy}{dx} =\frac{1}{\cos\left(\sin^{-1}(x)\right)}\end{gathered} \end{equation*}

Although correct, this formula is cumbersome to use, and can be simplified significantly with a bit of trigonometry. Let $\theta=\sin^{-1}(x)\text{,}$ so $\sin(\theta)=x\text{,}$ and construct a right angle triangle with angle $\theta\text{,}$ opposite side length $x$ and hypotenuse 1. The Pythagorean Theorem gives an adjacent side length of $\sqrt{1-x^2}\text{,}$ so $\cos\left(\sin^{-1}(x)\right)=\cos(\theta)=\sqrt{1-x^2}\text{.}$ Note that we choose the non-negative square root $\sqrt{1-x^2}$ since $\cos(\theta)\geq 0$ when $-\pi/2\leq\theta\leq\pi/2\text{.}$

Finally, the derivative of inverse sine is

\begin{equation*} \left(\sin^{-1}(x)\right)'=\frac{1}{\sqrt{1-x^2}} \end{equation*}
Example4.84. Derivative of Inverse Cosine.

Find the derivative of $\cos^{-1}(x)\text{.}$

Solution

Let $y=\cos^{-1}(x)\text{,}$ so $\cos(y)=x$ and $0\leq y\leq \pi\text{.}$ Next we differentiate implicitly:

\begin{equation*} \begin{gathered} \frac{d}{dx}\left(\cos y\right) =\frac{d}{dx}\left(x\right)\\ -\sin y\cdot\frac{dy}{dx} =1\\ \frac{dy}{dx} =-\frac{1}{\sin y}\end{gathered} \end{equation*}

This time, just for variety, we will leave the derivative in terms of $y$ and apply some trigonometry. Since $\cos y=x\text{,}$ we construct a triangle with angle $y\text{,}$ adjacent side length $x$ and hypotenuse $1\text{.}$ Solving for the opposite side length using Pythagorean Theorem we obtain $\sqrt{1-x^2}\text{.}$ Using this triangle we can see that $\sin y=\sqrt{1-x^2}$ ($0\leq y\leq \pi$). Substituting this into the equation for $dy/dx\text{,}$ we find that

\begin{equation*} \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\cos^{-1}(x)\right)=\frac{-1}{\sqrt{1-x^2}} \end{equation*}

In the following example we explore an alternate method of finding the derivative.

Example4.85. Derivative of Inverse Tangent.

Find the derivative of $\tan^{-1}(x)\text{.}$

Solution

We begin with $\tan\left(\tan^{-1}(x)\right)=x\text{.}$ Taking the derivative using the Chain Rule we obtain

\begin{equation*} \sec^2\left(\tan^{-1}(x)\right)\cdot\frac{d}{dx}\left(\tan^{-1}(x)\right)=1\text{,} \end{equation*}

which we rearrange to obtain

\begin{equation*} \frac{d}{dx}\left(\tan^{-1}(x)\right)=\frac{1}{\sec^2\left(\tan^{-1}(x)\right)}\text{.} \end{equation*}

Let $\tan^{-1}(x)=\theta\text{,}$ then $\tan(\theta)=x\text{.}$ We construct a triangle with angle $\theta\text{,}$ adjacent side $1$ and opposite side $x\text{.}$ The hypotenuse is $\sqrt{1+x^2}$ using Pythagorean Theorem. Then $\sec^2\left(\tan^{-1}(x)\right)=\sec^2(\theta)=\left(\sec(\theta)\right)^2=\left(\sqrt{1+x^2}\right)^2=1+x^2\text{.}$ Recall that $\sec(x)=1/\cos(x)\text{.}$ Finally, the derivative is

\begin{equation*} \frac{d}{dx}\left(\tan^{-1}(x)\right)=\frac{1}{1+x^2}\text{.} \end{equation*}

The derivatives of the three remaining inverse trigonometric functions can be found in a similar manner. The table below provides a summary of the derivatives of all six inverse trigonometric functions and their domains.

Example4.87. Inverse Functions and Chain Rule.

Find the derivative of $f(x)=\log_{2}\left(\sin^{-1}(x^{2}-3x)\right)\text{.}$

Solution
\begin{equation*} \begin{split} f'(x) \amp = \frac{1}{(\ln 2)\sin^{-1}(x^{2}-3x)}\frac{d}{dx}\left(\sin^{-1}(x^{2}-3x)\right) \\ \amp = \frac{1}{(\ln 2)\sin^{-1}(x^{2}-3x)}\left(\frac{1}{\sqrt{1-(x^{2}-3x)^{2}}} \frac{d}{dx}(x^{2}-3x)\right)\\ \amp = \frac{1}{(\ln 2)\sin^{-1}(x^{2}-3x)}\frac{1}{\sqrt{1-(x^{2}-3x)^{2}}} (2x-3) \\ \amp = \frac{2x-3}{(\ln 2)\sin^{-1}(x^{2}-3x)\sqrt{1-(x^{2}-3x)^{2}}} \end{split} \end{equation*}
Example4.88. Inverse Functions and Implicit Differentiation.

Find $\frac{dy}{dx}$ for $\cos^{-1}(xy) = x^{2}\text{.}$

Solution
\begin{equation*} \begin{split} \frac{d}{dx}\left(cos^{-1}(xy)\right) \amp = \frac{d}{dx}(x^{2}) \\ -\frac{1}{\sqrt{1-(xy)^{2}}}\frac{d}{dx}(xy) \amp = 2x \\ \frac{y + x \frac{dy}{dx}}{\sqrt{1-(xy)^{2}}} \amp = 2x \\ -\frac{x}{\sqrt{1-(xy)^{2}}}\frac{dy}{dx} \amp = 2x + \frac{y}{\sqrt{1-(xy)^{2}}} \\ \frac{dy}{dx} \amp = -\frac{\sqrt{1-(xy)^{2}}}{x}\left(2x + \frac{y}{\sqrt{1-(xy)^{2}}}\right) \end{split} \end{equation*}
Exercises for Section 4.8.

Find the derivative of the function.

1. $\ds f(x)=\csc^{-1} (5x^{2}+1)$

$\ds f'(x)= -\dfrac{10x}{(5x^{2}+1)\sqrt{(5x^{2}+1)^{2}-1}}$
Solution

\begin{aligned}\diff{f}{x} \amp = \diff{}{x} \csc^{-1}(5x^2+1)\\ \amp = -\frac{1}{|5x^2+1|\sqrt{(5x^2+1)^2-1}} \diff{}{x} (5x^2+1)\\ \amp = -\frac{10x}{|5x^2+1|\sqrt{(5x^2+1)^2-1}} \end{aligned}

2. $\ds f(x)=(\tan^{-1}(2x))^{3}$

$\ds f'(x)= \dfrac{6(\tan^{-1}(2x))^{2}}{1+4x^{2}}$
Solution

\begin{aligned}\diff{f}{x} \amp = \diff{}{x} \left(\tan^{-1}(2x)\right)^3\\ \amp = 3\left(\tan^{-1}(2x)\right)^2\diff{}{x} \tan^{-1}(2x)\\ \amp = 3\left(\tan^{-1}(2x)\right)^2 \cdot \frac{1}{1+4x^2} \diff{}{x} (2x)\\ \amp =3\left(\tan^{-1}(2x)\right)^2 \cdot \frac{1}{1+4x^2} \cdot 2 \\ \amp = \frac{6 \left(\tan^{-1}(2x)\right)^2}{1+4x^2} \end{aligned}

3. $\ds g(x)=\sqrt{e^{cos^{-1}(x)}}$

$\ds g'(x)=-\dfrac{\sqrt{e^{\cos^{-1}x}}}{2\sqrt{1-x^{2}}}$
Solution

\begin{aligned}\diff{g}{x} \amp = \diff{}{x} \sqrt{e^{\cos^{-1}(x)}} \\ \amp = \frac{1}{2\sqrt{e^{\cos^{-1}(x)}} } \diff{}{x} e^{\cos^{-1}(x)} \\ \amp = \frac{1}{2\sqrt{e^{\cos^{-1}(x)}} } \cdot e^{\cos^{-1}(x)} \diff{}{x} \cos^{-1}(x) \\ \amp = \frac{1}{2\sqrt{e^{\cos^{-1}(x)}} } \cdot e^{\cos^{-1}(x)} \cdot \frac{1}{\sqrt{1-x^2}} \\ \amp =\frac{\sqrt{e^{\cos^{-1}(x)}}} {2\sqrt{1-x^2}} \end{aligned}

4. $\ds f(t)=\ln(\sin^{-1}t)$

$\ds f'(t)= \dfrac{1}{\sqrt{1-t^{2}}\sin^{-1}t}$
Solution

\begin{aligned}\diff{f}{t} \amp = \diff{}{t} \ln\bigl(\sin^{-1} t\bigr)\\ \amp = \frac{1}{\sin^{-1}t} \diff{}{t} \sin^{-1}(t) \\ \amp = \frac{1}{\sin^{-1}t\sqrt{1-t^2}} \end{aligned}

5. $\ds f(x)=\sec^{-1}(x^{3/2})$

$\ds f'(x)= \dfrac{2}{2|x|\sqrt{x^{3}-1}}$
Solution

\begin{aligned}\diff{f}{x} \amp = \diff{}{x} \sec^{-1}(x^{3/2})\\ \amp = \frac{1}{|x^{3/2}| \sqrt{|x|^3-1}} \diff{}{x} x^{3/2} \\ \amp = \frac{3\sqrt{x}}{2|x^{3/2}| \sqrt{|x|^3-1}}\\ \amp = \frac{3}{2|x| \sqrt{|x|^3-1}} \end{aligned}

6. $\ds h(s)=\cos^{-1}(\log_{2}s)$

$\ds h'(s)= -\dfrac{1}{(\ln 2)s\sqrt{1-(\log_{2}s)^{2}}}$
Solution

\begin{aligned}\diff{h}{s} \amp = \diff{}{x} \cos^{-1}(\log_2 s)\\ \amp = -\frac{1}{\sqrt{1-(\log_2 s)^2}} \diff{}{s} \log_2 s\\ \amp = -\frac{1}{s\ln(2)\sqrt{1-(\log_2 s)^2}} \end{aligned}

7. $\ds f(x)=(\cot^{-1}x)^{1/3}$

$\ds f'(x)= -\dfrac{1}{3(1+x^{2})(\cot^{-1}x)^{2/3}}$
Solution

\begin{aligned}\diff{f}{x} \amp = \diff{}{x} (\cot^{-1}x)^{1/3} \\ \amp = \frac{1}{3} (\cot^{-1}x)^{-2/3}) \diff{}{x} \cot^{-1} x\\ \amp = \frac{-1}{3(1+x^2)} (\cot^{-1}x)^{-2/3}) \end{aligned}

8. $\ds g(t)=\sin^{-1}(3^{t})$

$\ds g'(t)= \dfrac{3^{t}(\ln 3)}{\sqrt{1-3^{2t}}}$
Solution

\begin{aligned}\diff{g}{t} \amp = \diff{}{t} \sin^{-1}(3^t) \\ \amp = \frac{1}{\sqrt{1-(3^t)^2}} \diff{}{t} 3^t \\ \amp = \frac{3^t \ln(3)}{\sqrt{1-3^{2t}}} \end{aligned}

Find $\frac{dy}{dx}$ by implicit differentiation.

1. $\ds \sin^{-1}(xy)+xy = x$

$\ds \frac{dy}{dx}= \left(1-y-\dfrac{y}{\sqrt{1-(xy)^{2}}}\right)\left(\dfrac{x}{\sqrt{1-(xy)^{2}}}+x\right)^{-1}$
Solution

\begin{aligned}\amp \diff{}{x}\left(\sin(xy)+xy\right) =\diff{}{x}(x) \amp \\ \amp \dfrac{1}{\sqrt{1-(xy)^{2}}}\left(y+xy'\right) + \left(y+xy'\right) = 1\amp \\ \amp \left(xy'\right) \left(\dfrac{1}{\sqrt{1-(xy)^{2}}} + 1\right) = 1 - y \left(\dfrac{1}{\sqrt{1-(xy)^{2}}} + 1\right)\amp \\\amp y' = \left(1-y-\dfrac{y}{\sqrt{1-(xy)^{2}}}\right)\left(\dfrac{x}{\sqrt{1-(xy)^{2}}} + x\right)^{-1}\amp \end{aligned}

2. $\ds \tan^{-1}(x-y)=xy$

$\ds \frac{dy}{dx} = \left(\dfrac{1}{1+(x-y)^{2}}-y\right)\left(\dfrac{1}{1+(x-y)^{2}}+x\right)^{-1}$
Solution

\begin{aligned}\amp \diff{}{x} \tan^{-1}(x-y) = \diff{}{x} xy\amp \\ \amp \frac{1}{1+(x-y)^2} \diff{}{x} (x-y) = xy' + y\amp \\ \amp \frac{1-y'}{1+(x-y)^2} = xy' + y\amp \\ \amp \frac{-y'}{1+(x-y)^2} - xy' = \frac{-1}{1+(x-y)^2} + y\amp \\ \amp y' = \frac{\frac{1}{1+(x-y)^2}-y}{\frac{1}{1+(x-y)^2}+x}\amp \end{aligned}

Given $f(x)=1+\ln(x-2)\text{,}$ first show that $f^{-1}$ exists, then compute $\left[f^{-1}\right]'(1)\text{.}$

$f^{-1}=1$
Solution

Let $y=f(x)=1+\ln(x-2)\text{.}$ Then,

\begin{equation*} \begin{split} y-1 \amp = \ln(x-2) \\ e^{y-2} + 2 \amp = x \end{split} \end{equation*}

Therefore $f^{-1}(x) = e^{x-1}+2\text{.}$

$f^{-1}(1) = e^{0}+2 = 3\text{.}$ So,

\begin{equation*} (f^{-1})'(1) = \frac{1}{x-2} \bigg\rvert_{x=3} = 1\text{.} \end{equation*}

The inverse cotangent function, denoted by $\cot^{-1}(x)\text{,}$ is defined to be the inverse of the restricted cotangent function: $\cot (x)\text{,}$ $0\lt x\lt \pi\text{.}$ Find the derivative of $\cot^{-1}(x)\text{.}$

$\ds\diff{}{y} \cot^{-1} y = \frac{-1}{1+y^2}$
Solution

We know that

\begin{equation*} \diff{}{x} \cot^{-1}(x) = \frac{-1}{1+x^2}\text{.} \end{equation*}

To derive this formula, let $y=\cot(x)\text{.}$ That is, $\cot^{-1} y = x\text{.}$ Then we know that

\begin{equation*} \diff{y}{x} = \diff{}{x} \frac{\cos x}{\sin x} = \frac{-1}{\sin^2 x}\text{.} \end{equation*}

We now take

\begin{equation*} \begin{split} \diff{}{x} \cot^{-1}y \amp = \diff{}{x} (x) \\ \diff{}{y} \cot^{-1} y \cdot \diff{y}{x} \amp = 1 \\ \diff{}{y} \cot^{-1} y \amp = -\sin^2 x \end{split} \end{equation*}

It remains to evaluate $\sin^2 x\text{.}$ We have $\cot(x)=y\text{,}$ and so we construct the followng right triangle $(0 \lt x \lt \pi)\text{:}$

Hence,

\begin{equation*} \sin x = \frac{1}{\sqrt{1+y^2}} \implies \sin^2 x = \frac{1}{1+y^2}\text{.} \end{equation*}

Altogether, we have shown that

\begin{equation*} \diff{}{y} \cot^{-1} y = -\sin^2 x = \frac{-1}{1+y^2}\text{.} \end{equation*}

This gives the desired formula.

The inverse secant function, denoted by $\sec^{-1}(x)\text{,}$ is defined to be the inverse of the restricted secant function: $\sec(x)\text{,}$ $x\in[0,\pi/2)\cup[\pi,3\pi/2)\text{.}$ Find the derivative of $\sec^{-1}(x)\text{.}$

$\ds\diff{}{x}\sec^{-1} x = \frac{1}{|x| \sqrt{x^2-1}}$
Solution

We know that the formula for the derivative of the inverse secant function is given by

\begin{equation*} \diff{}{x} \sec^{-1} x = \frac{1}{|x|\sqrt{x^2-1}}\text{.} \end{equation*}

To derive this formula, let

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

Differentiating both sides of this equation, we have

\begin{equation*} \frac{\sin(\sec^{-1} x)}{\cos^2(\sec^{-1} x)} \diff{}{x} \sec^{-1} x = 1\text{.} \end{equation*}

Rearranging, we have

\begin{equation*} \diff{}{x}\sec^{-1} x = \frac{\cos^2(\sec^{-1} x)}{\sin(\sec^{-1} x)}\text{.} \end{equation*}

Now, let $\theta = \sec^{-1} x\text{.}$ So we need to find $\cos^2 \theta$ and $\sin \theta\text{.}$ Given that $\sec \theta = x\text{,}$ we can construct the following right triangle (for $\theta \in [0,\pi/2) \cup [\pi,3\pi/2)$):

Hence,

\begin{equation*} \sin \theta = \frac{\sqrt{x^2-1}}{|x|}, \text{ and } \cos \theta = \frac{1}{x} \implies \cos^2\theta = \frac{1}{x^2}\text{.} \end{equation*}

Altogther, we have

\begin{equation*} \begin{split} \diff{}{x}\sec^{-1} x \amp = \frac{\cos^2(\theta)}{\sin(\theta)}\\ \amp = \frac{1}{x^2} \cdot \frac{|x|}{\sqrt{x^2-1}} \\ \amp = \frac{1}{|x| \sqrt{x^2-1}} \end{split} \end{equation*}

which agrees with the formula above.

The inverse cosecant function, denoted by $\csc^{-1}(x)\text{,}$ is defined to be the inverse of the restricted cosecant function: $\csc(x)\text{,}$ $x\in(0,\pi/2]\cup(\pi,3\pi/2]\text{.}$ Find the derivative of $\csc^{-1}(x)\text{.}$

$\ds\diff{}{x} \csc^{-1} x = -\frac{1}{|x|\sqrt{x^2-1}}$
Solution

We know that the formula for the derivative of the inverse cosecant function is

\begin{equation*} \diff{}{x} \csc^{-1} x = -\frac{1}{|x}\sqrt{x^2-1}\text{.} \end{equation*}

To derive this formula, we start with $\csc\bigl(\csc^{-1} x \bigr) = x\text{.}$ Then differentiating both sides, we find

\begin{equation*} -\frac{\cos(\csc^{-1}x)}{\sin^2(\csc^{-1} x)} \diff{}{x} \csc^{-1} x = 1\text{.} \end{equation*}

Therefore,

\begin{equation*} \diff{}{x} \csc^{-1} x = -\frac{\sin^2(\csc^{-1} x)}{\cos(\csc^{-1}x)}\text{.} \end{equation*}

Now, let $\theta = \csc^{-1} x\text{.}$ Then we need to find $\sin^2 \theta$ and $\cos \theta\text{.}$ Since $\csc(\theta) = x\text{,}$ we can construct the following right triangle (for $\theta \in [0,\pi/2) \cup [\pi,3\pi/2)$):

Hence,

\begin{equation*} \sin^2\theta = \frac{1}{x^2}, \text{ and } \cos\theta = \frac{\sqrt{x^2-1}}{|x|}\text{.} \end{equation*}

Altogether, we have found that

\begin{equation*} \begin{split} \diff{}{x} \csc^{-1} x \amp = -\frac{\sin^2(\csc^{-1} x)}{\cos(\csc^{-1}x)}\\ \amp = -\frac{\sin^2 \theta}{\cos\theta} \\ \amp = -\frac{|x|}{x^2\sqrt{x^2-1}}\\ \amp = -\frac{1}{|x|\sqrt{x^2-1}} \end{split} \end{equation*}

which agrees with the formula above.

Suppose $f(x)=x^3+4x+2\text{.}$ Find the slope of the tangent line to the graph of $g(x)=xf^{-1}(x)$ at the point where $x=7\text{.}$

50
Solution

Differentiating $g(x)\text{,}$ we have

\begin{equation*} g'(x) = f^{-1}(x) + x [f^{-1}]'(x) \implies g'(7) = f^{-1}(7) + 7 [f^{-1}]'(7)\text{.} \end{equation*}

Now notice that $f(1)=7\text{.}$ That is, $f^{-1}(7)=1\text{.}$ Since

\begin{equation*} f'(x) = 3x^2+4\text{,} \end{equation*}

we have

\begin{equation*} \begin{split} [f^{-1}]' (7) \amp = \frac{1}{f'[f^{-1}(7)]} \\ \amp = \frac{1}{f'(1)} \\ \amp = 3(1) + 4\\ \amp = 7 \end{split} \end{equation*}

Therefore,

\begin{equation*} g'(7) = (1) + (7)(7) = 50\text{.} \end{equation*}

Find the derivatives of $\sin^{-1}(x)+\cos^{-1}(x)$ and $(x^2+1)\tan^{-1}(x)\text{.}$

$\ds\diff{}{x} (x^2+1)\tan^{-1} x = 2x\tan^{-1} x + 1$
Solution

We first differentiate $\sin^{-1} x + \cos^{-1} x$ using the formulas for the derivatives of inverse trigonometric functions:

\begin{equation*} \begin{split} \diff{}{x} \left(\sin^{-1} x + \cos^{-1} x\right) \amp = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}}\\ \amp = 0 \end{split} \end{equation*}

Next, we differentiate $(x^2+1)\tan^{-1} x\text{:}$

\begin{equation*} \begin{split} \diff{}{x} (x^2+1)\tan^{-1} x \amp = 2x\tan^{-1} x + \frac{x^2+1}{1+x^2} \\ \amp = 2x\tan^{-1} x + 1 \end{split} \end{equation*}

Differentiate $y=\sin^{-1}(x^2)$ and $y=\tan^{-1}(3x)\text{.}$

$\ds\diff{}{x} \tan^{-1}(3x) = \frac{3}{1+9x^2}$
for $x \in (-1,1)$ and
for $x \in (-\infty,\infty)\text{.}$