Chapter 3.5, Problem 8E

To determine

To find: $\frac{dy}{dx}$ use the grapher to support your analysis if you are unsure to your answer.

Answer to Problem 8E

  $\frac{dy}{dx}=\frac{1+\cos x+x\sin x}{{(1+\cos x)}^2}$ .

Explanation of Solution

Given information:Given equation is $y=\frac{x}{1+\cos x}$ .

Calculation:Given equation is $y=\frac{x}{1+\cos x}$ .

  $\begin{array}{l}y=\frac{x}{1+\cos x}\\ \frac{dy}{dx}=\frac{(1+\cos x)\frac{d}{dx}(x)-x\frac{d}{dx}(1+\cos x)}{{(1+\cos x)}^2}\\ \frac{dy}{dx}=\frac{(1+\cos x)(1)-x(-\sin x)}{{(1+\cos x)}^2}\\ \frac{dy}{dx}=\frac{1+\cos x+x\sin x}{{(1+\cos x)}^2}\end{array}$

Graph is: