Chapter 4.1, Problem 5E

To determine

To calculate: The derivative of y with respect to x using Chain Rule.

Answer to Problem 5E

  $\frac{dy}{dx}=2\left(\frac{\sin x\times \left(\cos x+\cos 2x\right)}{{\left(1+\cos x\right)}^3}\right)$ .

Explanation of Solution

Given information: The equation of y and u is given as,

  $\begin{array}{l}y={\left(\frac{\sin x}{1+\cos x}\right)}^2\\ u=\frac{\sin x}{1+\cos x}\end{array}$

Formula used: The formula that is used here is Chain Rule which is as follows,

  $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$ ............ (1)

First calculate the derivative of u with respect to x, then substitute it in the equation of y and then apply the Chain Rule.

Calculation: Substitute u in y ,

  $y=u^2$

Apply the Chain Rule given in equation (1),

  $\begin{array}{l}\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\\ \frac{dy}{dx}=\frac{d\left(u^2\right)}{du}\times \frac{d\left(\frac{\sin x}{1+\cos x}\right)}{dx}\\ \frac{dy}{dx}=2u\times \left(\frac{1}{1+\cos x}\times \frac{d\sin x}{dx}+\sin x\times \frac{d\left(\frac{1}{1+\cos x}\right)}{dx}\right)\\ \frac{dy}{dx}=2u\times \left(\frac{\cos x}{1+\cos x}+\sin x\times \frac{1}{{\left(1+\cos x\right)}^2}\times \frac{d\left(1+\cos x\right)}{dx}\right)\\ \frac{dy}{dx}=2u\times \left(\frac{\cos x}{1+\cos x}+\sin x\times \frac{1}{{\left(1+\cos x\right)}^2}\times \left(0-\sin x\right)\right)\\ \frac{dy}{dx}=2u\times \left(\frac{\cos x}{1+\cos x}-\frac{{\sin }^{2}x}{{\left(1+\cos x\right)}^2}\right)\\ \frac{dy}{dx}=2u\times \left(\frac{\cos x\times \left(1+\cos x\right)-{\sin }^{2}x}{{\left(1+\cos x\right)}^2}\right)\\ \frac{dy}{dx}=2u\times \left(\frac{\cos x+{\cos }^{2}x-{\sin }^{2}x}{{\left(1+\cos x\right)}^2}\right)\\ \frac{dy}{dx}=2u\times \left(\frac{\cos x+\cos 2x}{{\left(1+\cos x\right)}^2}\right)\end{array}$

Substitute u,

  $\begin{array}{l}\frac{dy}{dx}=2\left(\frac{\sin x}{1+\cos x}\right)\times \left(\frac{\cos x+\cos 2x}{{\left(1+\cos x\right)}^2}\right)\\ \frac{dy}{dx}=2\left(\frac{\sin x\times \left(\cos x+\cos 2x\right)}{{\left(1+\cos x\right)}^3}\right)\end{array}$

The derivative of y with respect to x is $\frac{dy}{dx}=2\left(\frac{\sin x\times \left(\cos x+\cos 2x\right)}{{\left(1+\cos x\right)}^3}\right)$ .