Chapter 3, Problem 82EP

(a).

To determine

The differentiation of function.

Answer to Problem 82EP

The differentiation is $\frac{{\text{e}}^x-{\text{e}}^{-x}}{2}$

Explanation of Solution

Given:

  $y=\frac{e^x+e^{-x}}{2}$

Concept Used:

• The differentiation of $\frac{d}{dx}\left[e^x\right]=e^x$

Calculation:

Differentiation is as follow:

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

Conclusion:

  $\left[\frac{dy}{dx}\right]=\frac{e^x-e^{-x}}{2}$

(b).

To determine

The double derivative of the function

Answer to Problem 82EP

The double derivative is $\frac{e^x+e^{-x}}{2}$

Explanation of Solution

Given:

  $y=\frac{e^x+e^{-x}}{2}$

Concept Used:

• The differentiation of $\frac{d}{dx}\left[e^x\right]=e^x$

Calculation:

The first derivative from $\left(a\right)$ is $\left[\frac{dy}{dx}\right]=\frac{e^x-e^{-x}}{2}$

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

Conclusion:

  $\frac{d^{2}y}{d{x}^2}=\frac{e^x+e^{-x}}{2}$

(c).

To determine

To find the equation of the line tangent to the curve at $x=1$

Answer to Problem 82EP

The equation of the line tangent to the curve at $x=1$

  $y_{\text{tangent}}=\left(\frac{e^2+1}{2}\right)+\left(\frac{e^2-1}{2}\right)\left(x-1\right)$

Explanation of Solution

Given:

  $y=\frac{e^x+e^{-x}}{2}$

Concept Used:

tangent $=\frac{dy}{dx}$

Equation of tangent:

  $\begin{array}{l}y-y_1=m\left(x-x_1\right),\\ m=\text{slope}.\end{array}$

Calculation:

The first derivative from $\left(a\right)$ is $\left[\frac{dy}{dx}\right]=\frac{e^x-e^{-x}}{2}$

Slope of tangent is as follows:

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{x=1}={\left[\frac{e^1-e^{-1}}{2}\right]}_{x=1}\\ =\frac{e^2-1}{2e}\end{array}$

At $x=1$ ,the value of $y$ :

  $\begin{array}{c}y=\frac{e^1+e^{-1}}{2}\\ =\frac{e^2+1}{2e}\end{array}$

The value of $\left(x_1,y_1\right)=\left(1,\frac{e^2+1}{2e}\right)$

The equation of tangent:

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-\left(\frac{e^2+1}{2}\right)=\left(\frac{e^2-1}{2}\right)\left(x-1\right)\\ y=\left(\frac{e^2+1}{2}\right)+\left(\frac{e^2-1}{2}\right)\left(x-1\right)\end{array}$

Conclusion:

The equation of tangent is:

  $y_{\text{tangent}}=\left(\frac{e^2+1}{2}\right)+\left(\frac{e^2-1}{2}\right)\left(x-1\right)$

(d).

To determine

To find the equation of the line normal to the curve at $x=1$

Answer to Problem 82EP

The equation of the line normal to the curve at $x=1$ is $y_{normal}=\left(\frac{e^2+1}{2}\right)+\left(\frac{2e}{1+e^2}\right)\left(x-1\right)$

Explanation of Solution

Given:

  $y=\frac{e^x+e^{-x}}{2}$

Concept Used:

The normal is perpendicular to the tangent.

Slope of normal:

  $=\frac{-1}{m}$

Calculation:

The first derivative from $\left(a\right)$ is $\left[\frac{dy}{dx}\right]=\frac{e^x-e^{-x}}{2}$

And the slope is:

  $\begin{array}{c}\frac{-1}{m}=\frac{-1}{\frac{e^2-1}{2e}}\\ =\frac{-2e}{e^2-1}\\ =\frac{2e}{1+e^2}\end{array}$

The equation of normal is as follow:

  $\begin{array}{c}y-y_1=\frac{2e}{1+e^2}\left(x-x_1\right)\\ y-\left(\frac{e^2+1}{2}\right)=\left(\frac{2e}{1+e^2}\right)\left(x-1\right)\\ y=\left(\frac{e^2+1}{2}\right)+\left(\frac{2e}{1+e^2}\right)\left(x-1\right)\end{array}$

Conclusion:

  $y_{normal}=\left(\frac{e^2+1}{2}\right)+\left(\frac{2e}{1+e^2}\right)\left(x-1\right)$

(e).

To determine

To find any points where the tangent line is horizontal

Answer to Problem 82EP

Points where the tangent line is horizontal is $\left(0,1\right)$

Explanation of Solution

Given:

  $y=\frac{e^x+e^{-x}}{2}$

Concept Used:

The points at which the tangent line is horizontal, where the slope of the function is $0$ because a horizontal line's slope is $0$

Calculation:

The first derivative from $\left(a\right)$ is $\left[\frac{dy}{dx}\right]=\frac{e^x-e^{-x}}{2}$

  $\begin{array}{c}0=\frac{e^x-e^{-x}}{2}\\ 0=e^x-e^{-x}\\ 0=e^x-\frac{1}{e^x}\\ 0=\frac{e^{2x}-1}{e^x}\\ e^{2x}=1\\ 2x=0\\ x=0\end{array}$

The tangent line is horizontal when $x=0$

Plug in $x=0$ in the original function to find the $y$ value of the point

  $\begin{array}{c}y=\frac{e^0+e^0}{2}\\ =\frac{1+1}{2}\\ =\frac{2}{2}\\ =1\end{array}$

Conclusion:

The point at which the tangent line is horizontal is $\left(0,1\right)$