Chapter 3, Problem 2RCC

(a)

To determine

To State: The derivative of $y=x^n$.

Answer to Problem 2RCC

The derivative of $y=x^n$ is, $y^{\prime }=n{x}^{n-1}$.

Explanation of Solution

Binomial expansion:

${\left(x+h\right)}^n=x^n+n{x}^{n-1}h+\frac{n\left(n-1\right)}{2}{x}^{n-2}{h}^2+\mathrm{...}+nx{h}^{n-1}+h^n$

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^n-x^n}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(x^n+n{x}^{n-1}h+\frac{n\left(n-1\right)}{2}{x}^{n-2}{h}^2+\mathrm{...}+nx{h}^{n-1}+h^n\right)-x^n}{h}\\ =\underset{h\to 0}{\lim }\frac{n{x}^{n-1}h+\frac{n\left(n-1\right)}{2}{x}^{n-2}{h}^2+\mathrm{...}+nx{h}^{n-1}+h^n}{h}\\ =n{x}^{n-1}\end{array}$

Therefore, the derivative of y is $y^{\prime }=n{x}^{n-1}$.

(b)

To determine

To find: The derivative of $y=e^x$.

Answer to Problem 2RCC

The derivative of $y=e^x$ is, $y^{\prime }=e^x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}\\ =\underset{h\to 0}{\lim }\frac{e^x\left(e^h-1\right)}{h}\\ =e^x\underset{h\to 0}{\lim }\frac{\left(e^h-1\right)}{h}\end{array}$

Use $\underset{h\to 0}{\lim }\frac{\left(e^h-1\right)}{h}=1$, the derivative is $y^{\prime }=e^x$.

Therefore, the derivative of y is, $y^{\prime }=e^x$.

(c)

To determine

To find: The derivative of $y=b^x$.

Answer to Problem 2RCC

The derivative of y is, $y^{\prime }=b^x\ln b$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{b^{x+h}-b^x}{h}\\ =\underset{h\to 0}{\lim }\frac{b^x\left(b^h-1\right)}{h}\\ =b^x\underset{h\to 0}{\lim }\frac{\left(b^h-1\right)}{h}\end{array}$

Since $\underset{h\to 0}{\lim }\frac{\left(b^h-1\right)}{h}=\ln b$, the derivative is $y^{\prime }=b^x\ln b$.

Therefore, the derivative of y is $y^{\prime }=b^x\ln b$.

(d)

To determine

To find: The derivative of $y=\ln x$.

Answer to Problem 2RCC

The derivative of $y=\ln x$ is $y^{\prime }=\frac{1}{x}$.

Explanation of Solution

Result used:

The value $\underset{h\to 0}{\lim }{\left(1+t\right)}^{\frac{1}{t}}=e^t$

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\ln \left(x+h\right)-\ln x}{h}\\ =\underset{h\to 0}{\lim }\frac{\ln \left(\frac{x+h}{x}\right)}{h}\left(\ln \left(\frac{a}{b}\right)=\ln a-\ln b\right)\\ =\underset{h\to 0}{\lim }\frac{1}{h}\ln \left(1+\frac{h}{x}\right)\end{array}$

Use $\ln a^m=m\ln a$ in the above,

$y^{\prime }=\underset{h\to 0}{\lim }\ln {\left(1+\frac{h}{x}\right)}^{\frac{1}{h}}$

Let $t=\frac{h}{x}$ then t tends to zero as h tends to zero.

$\begin{array}{c}{y}^{\prime }=\underset{t\to 0}{\lim }\ln {\left(1+t\right)}^{\frac{1}{tx}}\\ =\underset{t\to 0}{\lim }{\left(\ln {\left(1+t\right)}^{\frac{1}{t}}\right)}^{\frac{1}{x}}\\ ={\left(\underset{t\to 0}{\lim }\ln {\left(1+t\right)}^{\frac{1}{t}}\right)}^{\frac{1}{x}}\end{array}$

Use the result stated above,

$\begin{array}{c}{y}^{\prime }=\ln e^{\frac{1}{x}}\\ =\frac{1}{x}\left(Q\ln e=1\right)\end{array}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{x}$.

(e)

To determine

To find: The derivative of $y={\log }_{b}x$.

Answer to Problem 2RCC

The derivative of $y={\log }_{b}x$ is, $y^{\prime }=\frac{1}{x\ln b}$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(y\right)\\ =\frac{d}{dx}\left({\log }_{b}x\right)\\ =\frac{1}{x\ln b}\left(Q\frac{d}{dx}\left({\log }_{a}x\right)=\frac{1}{x\ln a}\right)\end{array}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{x\ln b}$.

(f)

To determine

To find: The derivative of $y=\sin x$.

Answer to Problem 2RCC

The derivative of y is, $y^{\prime }=\cos x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin \left(x+h\right)-\sin x}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin x\cosh +\cos x\sinh -\sin x}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin x\left(\cosh -1\right)+\cos x\sinh }{h}\end{array}$

Simplify further,

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{\sin x\left(\cosh -1\right)}{h}+\underset{h\to 0}{\lim }\frac{\cos x\sinh }{h}\\ =\sin x\underset{h\to 0}{\lim }\frac{\left(\cosh -1\right)}{h}+\cos x\underset{h\to 0}{\lim }\frac{\sinh }{h}\\ =\sin x\left(0\right)+\cos x\left(1\right)\\ =\cos x\end{array}$

Therefore, the derivative of y is $y^{\prime }=\cos x$.

(g)

To determine

To find: The derivative of $y=\cos x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=-\sin x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos \left(x+h\right)-\cos x}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos x\cosh -\sin x\sinh -\cos x}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos x\left(\cosh -1\right)-\sin x\sinh }{h}\end{array}$

Simplify further,

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos x\left(\cosh -1\right)}{h}-\underset{h\to 0}{\lim }\frac{\sin x\sinh }{h}\\ =\cos x\underset{h\to 0}{\lim }\frac{\left(\cosh -1\right)}{h}-\sin x\underset{h\to 0}{\lim }\frac{\sinh }{h}\\ =\sin x\left(0\right)-\sin x\left(1\right)\\ =-\sin x\end{array}$

Therefore, the derivative of y is $y^{\prime }=-\sin x$.

(h)

To determine

To find: The derivative of $y=\tan x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }={\sec }^{2}x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\tan \left(x+h\right)-\tan x}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{\tan x+\tanh }{1-\tan x\tanh }-\tan x}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{\tan x+\tanh -\tan x+{\tan }^{2}x\tanh }{1-\tan x\tanh }}{h}\end{array}$

Simplify further,

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\left(\frac{\tanh }{h}\cdot \frac{1+{\tan }^{2}x}{1-\tan x\tanh }\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\frac{\sinh }{\cosh }}{h}\cdot \frac{{\sec }^{2}x}{1-\tan x\tanh }\right)\left(\begin{array}{l}Q1+{\tan }^{2}x={\sec }^{2}x\\ \tanh =\frac{\sinh }{\cosh }\end{array}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\sinh }{h}\cdot \frac{1}{\cosh }\cdot \frac{{\sec }^{2}x}{1-\tan x\tanh }\right)\\ ={\sec }^{2}x\end{array}$

Therefore, the derivative of y is $y^{\prime }={\sec }^{2}x$.

(i)

To determine

To find: The derivative of $y=\csc x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=-\csc x\cot x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\csc \left(x+h\right)-\csc x}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{\sin \left(x+h\right)}-\frac{1}{\sin x}}{h}\left(Q\csc x=\frac{1}{\sin x}\right)\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{\sin x\cosh +\cos x\sinh }-\frac{1}{\sin x}}{h}\end{array}$

Simplify further,

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{\sin x-\sin x\cosh -\cos x\sinh }{\left(\sin x\cosh +\cos x\sinh \right)\sin x}}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin x\left(1-\cosh \right)-\cos x\sinh }{h}\cdot \frac{1}{\sin \left(x+h\right)\sin x}\\ =\left[\sin x\left(\underset{h\to 0}{\lim }\frac{\left(1-\cosh \right)}{h}\right)-\left(\underset{h\to 0}{\lim }\frac{\cos x\sinh }{h}\right)\right]\cdot \left(\underset{h\to 0}{\lim }\frac{1}{\sin \left(x+h\right)\sin x}\right)\\ =\left[\sin x\left(0\right)-\cos x\left(1\right)\right]\frac{1}{{\sin }^{2}x}\end{array}$

That is, $y^{\prime }=\frac{-\cos x}{{\sin }^{2}x}$.

$y^{\prime }=-\csc x\cot x$

Therefore, the derivative of y is $y^{\prime }=-\csc x\cot x$.

(j)

To determine

To find: The derivative of $y=\sec x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\sec x\tan x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sec \left(x+h\right)-\sec x}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{\cos \left(x+h\right)}-\frac{1}{\cos x}}{h}\left(Q\sec x=\frac{1}{\cos x}\right)\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{\cos x\cosh -\sin x\sinh }-\frac{1}{\sin x}}{h}\end{array}$

Simplify further,

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{\cos x-\cos x\cosh +\sin x\sinh }{\left(\cos x\cosh -\sin x\sinh \right)\cos x}}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos x\left(1-\cosh \right)+\sin x\sinh }{h}\cdot \frac{1}{\cos \left(x+h\right)\cos x}\\ =\left[\cos x\left(\underset{h\to 0}{\lim }\frac{\left(1-\cosh \right)}{h}\right)+\left(\sin x\underset{h\to 0}{\lim }\frac{x\sinh }{h}\right)\right]\cdot \left(\underset{h\to 0}{\lim }\frac{1}{\cos \left(x+h\right)\cos x}\right)\\ =\left[\cos x\left(0\right)+\sin x\left(1\right)\right]\frac{1}{{\cos }^{2}x}\end{array}$

That is, $y^{\prime }=\frac{\sin x}{{\cos }^{2}x}$.

$y^{\prime }=\sec x\tan x$

Therefore, the derivative of y is, $y^{\prime }=\sec x\tan x$.

(k)

To determine

To find: The derivative of $y=\cot x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=-{\csc }^{2}x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\cot \left(x+h\right)-\cot x}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{\cos \left(x+h\right)}{\sin \left(x+h\right)}-\frac{\cos x}{\sin x}}{h}\left(Q\sec x=\frac{1}{\cos x}\right)\\ =\underset{h\to 0}{\lim }\frac{\sin x\cos \left(x+h\right)-\cos x\sin \left(x+h\right)}{h\sin \left(x+h\right)\sin x}\end{array}$

Simplify further,

$\begin{array}{l}{y}^{\prime }=\underset{h\to 0}{\lim }\frac{\sin x\left[\cos x\cosh -\sin x\sinh \right]-\cos x\left[\sin x\cosh +\cos x\sinh \right]}{h\sin \left(x+h\right)\sin x}\\ =\underset{h\to 0}{\lim }\frac{-\sin x{\sin }^{2}h-{\cos }^{2}x\sinh }{h\sin \left(x+h\right)\sin x}\\ =\underset{h\to 0}{\lim }\frac{-\left({\sin }^{2}x+{\cos }^{2}x\right)\sinh }{h\sin \left(x+h\right)\sin x}\\ =\underset{h\to 0}{\lim }\frac{-\sinh }{h\sin \left(x+h\right)\sin x}\left(Q{\sin }^{2}x+{\cos }^{2}x=1\right)\end{array}$

Apply the limit,

$\begin{array}{c}{y}^{\prime }=\underset{h\to 0}{\lim }\left(\frac{-\sinh }{h}\cdot \frac{1}{\sin \left(x+h\right)\sin x}\right)\\ =-\left(1\right)\frac{1}{\sin x\sin x}\\ =-\frac{1}{{\sin }^{2}x}\\ =-{\csc }^{2}x\end{array}$

Therefore, the derivative of y is $y^{\prime }=-{\csc }^{2}x$.

(l)

To determine

To find: The derivative of $y={\sin }^{-1}x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\frac{1}{\sqrt{1-x^2}}$.

Explanation of Solution

Calculation:

Consider the function $y={\sin }^{-1}x$.

That is $\sin y=x\text{ and }-\frac{\pi }{2}\le y\le \frac{\pi }{2}$

Differentiate implicitly with respect x,

$\begin{array}{l}\frac{d}{dx}\sin y=\frac{d}{dx}x\\ \cos y\frac{dy}{dx}=1\end{array}$

$\frac{dy}{dx}=\frac{1}{\cos y}$ (1)

Since $\cos y\ge 0\text{ and }-\frac{\pi }{2}\le y\le \frac{\pi }{2}$.

$\begin{array}{l}{\cos }^{2}y=1-{\sin }^{2}y\\ \cos y=\sqrt{1-{\sin }^{2}y}\\ \cos y=\sqrt{1-x^2}\left(Q\sin y=x\right)\end{array}$

Substitute $\cos y=\sqrt{1-x^2}$ in equation (1),

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{\sqrt{1-x^2}}$.

(m)

To determine

To find: The derivative of $y={\cos }^{-1}x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\frac{-1}{\sqrt{1-x^2}}$.

Explanation of Solution

Calculation:

Consider the function $y={\cos }^{-1}x$.

That is $\cos y=x\text{ and }\pi \le y\le 2\pi$

Differentiate implicitly with respect x,

$\begin{array}{l}\frac{d}{dx}\cos y=\frac{d}{dx}x\\ -\sin y\frac{dy}{dx}=1\end{array}$

$\frac{dy}{dx}=-\frac{1}{\sin y}$ (1)

Since $\sin y\ge 0\text{ and }\pi \le y\le 2\pi$.

$\begin{array}{l}{\sin }^{2}y=1-{\cos }^{2}y\\ \sin y=\sqrt{1-{\cos }^{2}y}\\ \sin y=\sqrt{1-x^2}\left(Q\cos y=x\right)\end{array}$

Substitute $\sin y=\sqrt{1-x^2}$ in equation (1),

$\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}$

Therefore, the derivative of y is $y^{\prime }=\frac{-1}{\sqrt{1-x^2}}$.

(n)

To determine

To find: The derivative of $y={\tan }^{-1}x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\frac{1}{1+x^2}$.

Explanation of Solution

Calculation:

Consider the function $y={\tan }^{-1}x$.

That is, $\tan y=x$

Differentiate implicitly with respect x,

$\begin{array}{l}\frac{d}{dx}\tan y=\frac{d}{dx}x\\ {\sec }^{2}y\frac{dy}{dx}=1\\ \frac{dy}{dx}=\frac{1}{{\sec }^{2}y}\\ \frac{dy}{dx}=\frac{1}{1+{\tan }^{2}y}\\ \frac{dy}{dx}=\frac{1}{1+x^2}\left(Q\tan y=x\right)\end{array}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{1+x^2}$.

(o)

To determine

To find: The derivative of $y=\sinh x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\cosh x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(\sinh x\right)\\ =\frac{d}{dx}\left(\frac{e^x-e^{-x}}{2}\right)\left(Q\sinh x=\frac{e^x-e^{-x}}{2}\right)\\ =\frac{1}{2}\left(e^x+e^{-x}\right)\\ =\cosh x\end{array}$

Therefore, the derivative of y is $y^{\prime }=\cosh x$.

(p)

To determine

To find: The derivative of $y=\cosh x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\sinh x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(\cosh x\right)\\ =\frac{d}{dx}\left(\frac{e^x+e^{-x}}{2}\right)\left(Q\cosh x=\frac{e^x+e^{-x}}{2}\right)\\ =\frac{1}{2}\left(e^x-e^{-x}\right)\\ =\sinh x\end{array}$

Therefore, the derivative of y is $y^{\prime }=\sinh x$.

(q)

To determine

To find: The derivative of $y=\tanh x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }={\mathrm{sech}}^{2}x$.

Explanation of Solution

Calculation:

Obtain the derivative of y.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(\tanh x\right)\\ =\frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right)\\ =\frac{\cosh x\frac{d}{dx}\left(\sinh x\right)-\sinh x\frac{d}{dx}\left(\cosh x\right)}{{\cosh }^{2}x}\\ =\frac{\cosh x\left(\cosh x\right)-\sinh x\left(\sinh x\right)}{{\cosh }^{2}x}\end{array}$

Simplify further,

$\begin{array}{l}{y}^{\prime }=\frac{{\cosh }^{2}x-{\sinh }^{2}x}{{\cosh }^{2}x}\\ =\frac{1}{{\cosh }^{2}x}\left(Q{\cosh }^{2}x-{\sinh }^{2}x=1\right)\\ ={\mathrm{sech}}^{2}x\left(Q\frac{1}{\cosh x}=\mathrm{sech}x\right)\end{array}$

Therefore, the derivative of y is $y^{\prime }={\mathrm{sech}}^{2}x$.

(r)

To determine

To find: The derivative of $y={\sinh }^{-1}x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\frac{1}{\sqrt{x^2+1}}$.

Explanation of Solution

Calculation:

Consider the function $y={\sinh }^{-1}x$

That is, $\sinh y=x$.

$\begin{array}{l}x=\frac{e^y-e^{-y}}{2}\left(Q\sinh x=\frac{e^x-e^{-x}}{2}\right)\\ x=\frac{e^y-e^{-y}}{2}\left(\frac{e^y}{e^y}\right)\\ x=\frac{e^{2y}-1}{2e^y}\\ 2x{e}^y=e^{2y}-1\end{array}$

Simplify further,

$\begin{array}{l}{e}^{2y}-2x{e}^y-1=0\\ {\left(e^y\right)}^2-2x{e}^y-1=0\end{array}$

Solve the above quadric equation,

$\begin{array}{c}{e}^y=\frac{2x+\sqrt{{\left(2x\right)}^2-4\left(1\right)\left(-1\right)}}{2}\\ =\frac{2x+\sqrt{4x^2+4}}{2}\\ =\frac{2x+2\sqrt{x^2+1}}{2}\\ =x+\sqrt{x^2+1}\end{array}$

That is ${\sinh }^{-1}x=x+\sqrt{x^2+1}$

Obtain the derivative $y={\sinh }^{-1}x$.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left({\sinh }^{-1}x\right)\\ =\frac{d}{dx}\left(\ln \left(x+\sqrt{x^2+1}\right)\right)\\ =\frac{1}{x+\sqrt{x^2+1}}\frac{d}{dx}\left(x+\sqrt{x^2+1}\right)\\ =\frac{1}{x+\sqrt{x^2+1}}\left(1+\frac{1}{2}{\left(x^2+1\right)}^{\frac{1}{2}-1}\left(2x\right)\right)\end{array}$

Simplify in terms and obtain the derivative,

$\begin{array}{c}{y}^{\prime }=\frac{1}{x+\sqrt{x^2+1}}\left(1+x{\left(x^2+1\right)}^{-\frac{1}{2}}\right)\\ =\frac{1}{x+\sqrt{x^2+1}}\left(1+\frac{x}{{\left(x^2+1\right)}^{\frac{1}{2}}}\right)\\ =\frac{1}{x+\sqrt{x^2+1}}\left(\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\right)\\ =\frac{1}{\sqrt{x^2+1}}\end{array}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{\sqrt{x^2+1}}$.

(s)

To determine

To find: The derivative of $y={\cosh }^{-1}x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\frac{1}{\sqrt{x^2-1}}$.

Explanation of Solution

Calculation:

Consider the function $y={\cosh }^{-1}x$

That is, $\cosh y=x$.

$\begin{array}{l}x=\frac{e^y-e^{-y}}{2}\left(Q\cosh x=\frac{e^x+e^{-x}}{2}\right)\\ x=\frac{e^y+e^{-y}}{2}\left(\frac{e^y}{e^y}\right)\\ x=\frac{e^{2y}+1}{2e^y}\\ 2x{e}^y=e^{2y}+1\end{array}$

Simplify further,

$\begin{array}{l}{e}^{2y}-2x{e}^y+1=0\\ {\left(e^y\right)}^2-2x{e}^y+1=0\end{array}$

Solve the above quadric equation,

$\begin{array}{c}{e}^y=\frac{2x+\sqrt{{\left(2x\right)}^2-4\left(1\right)\left(1\right)}}{2}\\ =\frac{2x+\sqrt{4x^2-4}}{2}\\ =\frac{2x+2\sqrt{x^2-1}}{2}\\ =x+\sqrt{x^2-1}\end{array}$

That is, ${\cosh }^{-1}x=x+\sqrt{x^2-1}$.

Obtain the derivative $y={\cosh }^{-1}x$.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left({\cosh }^{-1}x\right)\\ =\frac{d}{dx}\left(\ln \left(x+\sqrt{x^2-1}\right)\right)\\ =\frac{1}{x+\sqrt{x^2-1}}\frac{d}{dx}\left(x+\sqrt{x^2-1}\right)\\ =\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}{\left(x^2-1\right)}^{\frac{1}{2}-1}\left(2x\right)\right)\end{array}$

Simplify in terms and obtain the derivative,

$\begin{array}{c}{y}^{\prime }=\frac{1}{x+\sqrt{x^2-1}}\left(1+x{\left(x^2-1\right)}^{-\frac{1}{2}}\right)\\ =\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{x}{{\left(x^2-1\right)}^{\frac{1}{2}}}\right)\\ =\frac{1}{x+\sqrt{x^2-1}}\left(\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\right)\\ =\frac{1}{\sqrt{x^2-1}}\end{array}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{\sqrt{x^2-1}}$.

(t)

To determine

To find: The derivative of $y={\tanh }^{-1}x$.

Answer to Problem 2RCC

The derivative of y is $y^{\prime }=\frac{1}{1-x^2}$.

Explanation of Solution

Calculation:

Consider the function $y={\tanh }^{-1}x$

That is, $\tanh y=x$ .

$\begin{array}{l}x=\frac{e^y-e^{-y}}{e^y+e^{-y}}\left(Q\tanh x=\frac{e^x+e^{-x}}{e^x+e^{-x}}\right)\\ x=\frac{e^y-e^{-y}}{e^y+e^{-y}}\left(\frac{e^y}{e^y}\right)\\ x=\frac{e^{2y}-1}{e^{2y}+1}\\ x\left(e^{2y}+1\right)=e^{2y}-1\end{array}$

Simplify further,

$\begin{array}{l}x\left(e^{2y}+1\right)-e^{2y}+1=0\\ \left(x-1\right)e^{2y}+\left(x+1\right)=0\\ e^{2y}=\frac{-\left(x+1\right)}{\left(x-1\right)}\end{array}$

Take natural logarithm on both sides,

$\begin{array}{l}\ln e^{2y}=\ln \frac{-\left(x+1\right)}{\left(x-1\right)}\\ 2y=\ln \left(x+1\right)-\ln \left[-\left(x-1\right)\right]\\ y=\frac{1}{2}\left[\ln \left(x+1\right)-\ln \left(1-x\right)\right]\end{array}$

That is, ${\tanh }^{-1}x=\frac{1}{2}\left[\ln \left(x+1\right)-\ln \left(1-x\right)\right]$ .

Obtain the derivative $y={\cosh }^{-1}x$.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left({\tanh }^{-1}x\right)\\ =\frac{d}{dx}\left(\frac{1}{2}\left[\ln \left(x+1\right)-\ln \left(1-x\right)\right]\right)\\ =\frac{1}{2}\left[\frac{1}{x+1}-\frac{1}{1-x}\left(-1\right)\right]\\ =\frac{1}{2}\left[\frac{1-x+x+1}{1-x^2}\right]\\ =\frac{1}{1-x^2}\end{array}$

Therefore, the derivative of y is $y^{\prime }=\frac{1}{1-x^2}$.