Chapter 3, Problem 39RE

To determine

To calculate: Find $\frac{d^{2}y}{d{x}^2}$ .

Answer to Problem 39RE

  $\text{The value is }\frac{d^{2}y}{d{x}^2}=\frac{1+{\sin }^{2}x}{{\cos }^{3}x}$ .

Explanation of Solution

Given Information: The function is $y=\frac{1}{\cos x}$ .

Calculation:

The given function is,

  $y=\frac{1}{\cos x}$

Find $\frac{dy}{dx}$ as follows:

  $\begin{array}{c}y=\frac{1}{\cos x}\\ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{\cos x}\right)\left(\text{Substitute the value }y\text{ in }\frac{dy}{dx}\right)\\ =\frac{\left(\cos x\right)\frac{d}{dx}\left(1\right)-1\frac{d}{dx}\left(\cos x\right)}{{\left(\cos x\right)}^2}\left(\text{Apply Quotient rule}\right)\\ =\frac{\cos x\times 0-1\times \left(-\sin x\right)}{{\cos }^{2}x}\left(\frac{d}{dx}\left(\cos x\right)=-\sin x,\frac{d}{dx}\left(1\right)=0\right)\\ =\frac{0+\sin x}{{\cos }^{2}x}\left(\text{Simplify}\right)\\ =\frac{\sin x}{{\cos }^{2}x}\end{array}$

Find $\frac{d^{2}y}{d{x}^2}$ as follows:

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left(\frac{\sin x}{{\cos }^{2}x}\right)\left(\text{Substitute the value of }\frac{dy}{dx}\right)\\ =\frac{\left({\cos }^{2}x\right)\frac{d}{dx}\left(\sin x\right)-\left(\sin x\right)\frac{d}{dx}\left({\cos }^{2}x\right)}{{\left({\cos }^{2}x\right)}^2}\left(\text{Apply Quotient rule}\right)\\ =\frac{\left({\cos }^{2}x\times \cos x\right)-\left(\sin x\right)\left(-2\cos x\sin x\right)}{{\cos }^{4}x}\left(\frac{d}{dx}\left(\sin x\right)=\cos x,\frac{d}{dx}\left({\cos }^{2}x\right)=-2\cos x\sin x\right)\\ =\frac{{\cos }^{3}x+2{\sin }^{2}x\cos x}{{\cos }^{4}x}\left(\text{Simplify}\right)\\ =\frac{{\cos }^{2}x+2{\sin }^{2}x}{{\cos }^{3}x}\\ =\frac{1+{\sin }^{2}x}{{\cos }^{3}x}\left({\cos }^{2}x+{\sin }^{2}x=1\right)\end{array}$

Conclusion:

  $\text{The value is }\frac{d^{2}y}{d{x}^2}=\frac{1+{\sin }^{2}x}{{\cos }^{3}x}$ .