Chapter 3, Problem 38RE

To determine

To calculate: The find the slope of the curve at $x=\pi$ .

Answer to Problem 38RE

  $\text{The slope of the function} y=\frac{x}{x+\sin x}\text{ is }\frac{1}{\pi }$ .

Explanation of Solution

Given Information: The function is $y=\frac{x}{x+\sin x}$ .

Calculation:

The given function is,

  $y=\frac{x}{x+\sin x}$

The slope of a function is the rate of change of the function. That is, it is equal to the derivative of the function at x as follows:

  $\begin{array}{c}y=\frac{x}{x+\sin x}\\ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{x+\sin x}\right)\left(\text{Substitute the value }y\text{ in }\frac{dy}{dx}\right)\\ =\frac{\left(x+\sin x\right)\frac{d}{dx}\left(x\right)-x\frac{d}{dx}\left(x+\sin x\right)}{{\left(x+\sin x\right)}^2}\left(\text{Apply Quotient rule}\right)\\ =\frac{\left(x+\sin x\right)\times 1-x\times \left(1+\cos x\right)}{{\left(x+\sin x\right)}^2}\left(\frac{d}{dx}\left(x+\sin x\right)=1+\cos x,\frac{d}{dx}\left(x\right)=1\right)\\ =\frac{\left(x+\sin x\right)-\left(x+x\cos x\right)}{{\left(x+\sin x\right)}^2}\left(\text{Simplify}\right)\\ =\frac{x+\sin x-x-x\cos x}{{\left(x+\sin x\right)}^2}\\ =\frac{\sin x-x\cos x}{{\left(x+\sin x\right)}^2}\end{array}$

The slope at the point $x=\pi$ is,

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{x=\pi }={\left[\frac{\sin x-x\cos x}{{\left(x+\sin x\right)}^2}\right]}_{x=\pi }\\ =\frac{\sin \pi -\pi \cos \pi }{{\left(\pi +\sin \pi \right)}^2}\\ =\frac{0-\pi \left(-1\right)}{{\left(\pi +0\right)}^2}\left(\cos \pi =-1,\sin \pi =0\right)\\ =\frac{\pi }{{\pi }^2}\\ =\frac{1}{\pi }\end{array}$

Conclusion:

  $\text{The slope of the function} y=\frac{x}{x+\sin x}\text{ is }\frac{1}{\pi }$ .