Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

Find the derivative of the function.

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

   Answer
   \(\ds f'(x)= -\dfrac{10x}{(5x^{2}+1)\sqrt{(5x^{2}+1)^{2}-1}}\)
   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. \(\ds f(x)=(\tan^{-1}(2x))^{3}\)

   Answer
   \(\ds f'(x)= \dfrac{6(\tan^{-1}(2x))^{2}}{1+4x^{2}}\)
   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. \(\ds g(x)=\sqrt{e^{cos^{-1}(x)}}\)

   Answer
   \(\ds g'(x)=-\dfrac{\sqrt{e^{\cos^{-1}x}}}{2\sqrt{1-x^{2}}}\)
   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. \(\ds f(t)=\ln(\sin^{-1}t)\)

   Answer
   \(\ds f'(t)= \dfrac{1}{\sqrt{1-t^{2}}\sin^{-1}t}\)
   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. \(\ds f(x)=\sec^{-1}(x^{3/2})\)

   Answer
   \(\ds f'(x)= \dfrac{2}{2|x|\sqrt{x^{3}-1}}\)
   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. \(\ds h(s)=\cos^{-1}(\log_{2}s)\)

   Answer
   \(\ds h'(s)= -\dfrac{1}{(\ln 2)s\sqrt{1-(\log_{2}s)^{2}}}\)
   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. \(\ds f(x)=(\cot^{-1}x)^{1/3}\)

   Answer
   \(\ds f'(x)= -\dfrac{1}{3(1+x^{2})(\cot^{-1}x)^{2/3}}\)
   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. \(\ds g(t)=\sin^{-1}(3^{t})\)

   Answer
   \(\ds g'(t)= \dfrac{3^{t}(\ln 3)}{\sqrt{1-3^{2t}}}\)
   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. \(\ds \sin^{-1}(xy)+xy = x\)

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

   Answer
   \(\ds \frac{dy}{dx} = \left(\dfrac{1}{1+(x-y)^{2}}-y\right)\left(\dfrac{1}{1+(x-y)^{2}}+x\right)^{-1}\)
   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 \(\left[f^{-1}\right]'(1)\text{.}\)

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 \(\cot^{-1}(x)\text{.}\)

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 \(\sec^{-1}(x)\text{.}\)

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 \(\csc^{-1}(x)\text{.}\)

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 \(x=7\text{.}\)

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

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