Chapter 3.4, Problem 45E

(a)

To determine

To find: The equation of the tangent line to the curve at the point.

Answer to Problem 45E

The equation of the tangent line to the curve $y=\frac{2}{\left(1+e^{-x}\right)}$ at $\left(0,1\right)$ is $y=\frac{x}{2}+1$.

Explanation of Solution

Given:

The function is $y=\frac{2}{\left(1+e^{-x}\right)}$.

Derivative Rule: Quotient Rule

If $f\left(x\right)$. and $g\left(x\right)$ are both differentiable function, then

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^2}$ (1)

Formula used:

The equation of the tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$ (2)

where, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1}$.

Calculation:

The derivative of $y$ is $\frac{dy}{dx}$, which is obtained as follows,

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

Apply the quotient rule as shown in equation (1),

$\begin{array}{c}\frac{d}{dx}\left[\frac{2}{\left(1+e^{-x}\right)}\right]=\frac{\left(1+e^{-x}\right)\frac{d}{dx}\left[2\right]-2\frac{d}{dx}\left[\left(1+e^{-x}\right)\right]}{{\left(1+e^{-x}\right)}^2}\\ =\frac{\left(1+e^{-x}\right)\left[0\right]-2\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(e^{-x}\right)\right)}{{\left(1+e^{-x}\right)}^2}\\ =\frac{0-2\left(0+\left(-e^{-x}\right)\right)}{{\left(1+e^{-x}\right)}^2}\\ =\frac{2\left(e^{-x}\right)}{{\left(1+e^{-x}\right)}^2}\end{array}$

Therefore, the derivative of $y=\frac{2}{\left(1+e^{-x}\right)}$ is $\frac{dy}{dx}\underset{\_}{=\frac{2\left(e^{-x}\right)}{{\left(1+e^{-x}\right)}^2}}$.

The slope of the tangent line at $\left(0,1\right)$ is computed as follows,

$\begin{array}{c}m={\frac{dy}{dx}|}_{x=0}\\ \frac{2\left(e^0\right)}{{\left(1+e^0\right)}^2}\\ =\frac{2\left(1\right)}{{\left(2\right)}^2}\left[Q{e}^0=1\right]\\ =\frac{1}{2}\end{array}$

Thus, the slope of the tangent line at (0,1) is $m=\frac{1}{2}$.

Substitute $\left(0,1\right)$ for $\left(x_1,y_1\right)$ and $m=\frac{1}{2}$ in equation (1),

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

Therefore, the equation of the tangent line to the curve $y=\frac{2}{\left(1+e^{-x}\right)}$ at $\left(0,1\right)$ is $y=\frac{x}{2}+1$.

(b)

To determine

To sketch: The graph of the curve and the tangent line.

Explanation of Solution

Given:

The equation of the curve is $y=\frac{2}{\left(1+e^{-x}\right)}$.

The equation of the tangent line is $y=\frac{x}{2}+1$.

Graph:

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

From the Figure 1, it is observed that the equation of the tangent line touches on the curve $y=\frac{2}{\left(1+e^{-x}\right)}$ at the point $\left(0,1\right)$.