Chapter 3.7, Problem 28E

To determine

To calculate: The equation of tangent line to the curve $y=\frac{\ln x}{x}$ at the points $\left(1,0\right)\text{ and }\left(e,\frac{1}{e}\right)$ . Also plot the function along with its tangent lines.

Answer to Problem 28E

The equation of tangent line to the curve $y=\frac{\ln x}{x}$ at the point $\left(e,\frac{1}{e}\right)$ is $y=\frac{1}{e}$ and the point $\left(1,0\right)$ the equation of tangent is $y=x-1$ .The blue curve depicts the graph of $y=x-1$ , green color depicts the graph of $y=\frac{1}{e}$ and red curve depicts the graph of $y=\frac{\ln x}{x}$ .

Explanation of Solution

Given information:

The equation of curve $y=\frac{\ln x}{x}$ .

Formula used:

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

The point slope form of an equation is $\left(y-y_0\right)=m\left(x-x_0\right)$ , where m is the slope, $\left(x_0,y_0\right)$ is the point through which the curve pass.

Calculation:

Consider the equation of curve $y=\frac{\ln x}{x}$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\frac{\ln x}{x}\right)\\ y\text{'}=\frac{d}{dx}\left(\frac{\ln x}{x}\right)\end{array}$

Recall that quotient rule for differentiation is $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}$ where f and g are functions of x .

Apply it.

  $\begin{array}{c}y\text{'}=\frac{d}{dx}\left(\frac{\ln x}{x}\right)\\ y\text{'}\left(x\right)=\frac{x\left(\ln x\right)\text{'}-\ln x\left(x\right)\text{'}}{x^2}\\ =\frac{x\frac{1}{x}-\ln x}{x^2}\\ =\frac{1-\ln x}{x^2}\end{array}$

Recall that the point slope form of an equation is $\left(y-y_0\right)=m\left(x-x_0\right)$ , where m is the slope, $\left(x_0,y_0\right)$ is the point through which the curve pass.

It is provided that the curve passes through the point $\left(1,0\right)$ , so $\left(1,0\right)=\left(x_0,y_0\right)$ . Also slope is derivative of the curve at the point $\left(1,0\right)$ .

Substitute x as 1, in the derivative found above,

  $\begin{array}{c}y\text{'}\left(1\right)=\frac{1-\ln \left(1\right)}{{\left(1\right)}^2}\\ =\frac{1-0}{1}\\ =1\end{array}$

Therefore, slope is $m=1$ .

Now, the equation of tangent line to the curve $y=\frac{\ln x}{x}$ is,

  $\begin{array}{c}\left(y-y_0\right)=m\left(x-x_0\right)\\ y-0=1\left(x-1\right)\\ y=x-1\end{array}$

Thus, the equation of tangent line to the curve $y=\frac{\ln x}{x}$ at the point $\left(1,0\right)$ is $y=x-1$ .

It is provided that the curve passes through the point $\left(e,\frac{1}{e}\right)$ , so $\left(e,\frac{1}{e}\right)=\left(x_0,y_0\right)$ . Also slope is derivative of the curve at the point $\left(e,\frac{1}{e}\right)$ .

Substitute xas $e$ , in the derivative found above,

  $\begin{array}{c}y\text{'}\left(1\right)=\frac{1-\ln \left(e\right)}{{\left(e\right)}^2}\\ =\frac{1-1}{e^2}\\ =0\end{array}$

Therefore, slope is $m=0$ .

Now, the equation of tangent line to the curve $y=\frac{\ln x}{x}$ is,

  $\begin{array}{c}\left(y-y_0\right)=m\left(x-x_0\right)\\ y-\frac{1}{e}=0\left(x-e\right)\\ y=\frac{1}{e}\end{array}$

Thus, the equation of tangent line to the curve $y=\frac{\ln x}{x}$ at the point $\left(e,\frac{1}{e}\right)$ is $y=\frac{1}{e}$ .

Now, use a graphing window to plot the function and its tangent lines. Sketch the graph of $y=\frac{\ln x}{x}$ , $y=x-1$ and $y=\frac{1}{e}$ .

The blue curve depicts the graph of $y=x-1$ , green color depicts the graph of $y=\frac{1}{e}$ and red curve depicts the graph of $y=\frac{\ln x}{x}$ .

Observe that the both the lines $y=\frac{1}{e}$ and $y=x-1$ are a tangent to the curve $y=\frac{\ln x}{x}$ and touch the curve at only one point.

The line $y=\frac{1}{e}$ touch the curve at the point $\left(e,\frac{1}{e}\right)$ and the line $y=x-1$ touch the curve at the point $\left(1,0\right)$ .