Chapter 2.5, Problem 10E

To determine

To calculate: The derivative $\frac{dy}{dx}$ of the function

Answer to Problem 10E

The derivative of the function is. $-\frac{1}{1+\sin x}$ .

Explanation of Solution

Given information:

The function is $y=\frac{\cos x}{1+\sin x}$ .

Concept used:

The product rule is $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^2}$ and the formulae used are $\frac{d}{dx}\left(\sin x\right)=\cos x$ , $\frac{d}{dx}\left(\cos x\right)=-\sin x$ ,also ${\sin }^{2}x+{\cos }^{2}x=1$

Calculation:

The function is $y=\frac{\cos x}{1+\sin x}$ .

Differentiate the function with respect to $x$ .

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

Solve further.

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

Conclusion: The derivative of the function is $-\frac{1}{1+\sin x}$ .