Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises: 11

Answer

f'($\theta$)=sec($\theta$)($tan^{2}$($\theta$)+$sec^{2}$($\theta$))

Work Step by Step

The product rule states that if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x). If f($\theta$)=tan($\theta$)sec($\theta$), then we can set h(x)=tan($\theta$) and g(x)=sec($\theta$), and compute the derivative of the function f($\theta$). f'($\theta$)=tan($\theta$)$\frac{d}{d\theta}$[sec($\theta$)]+sec($\theta$)$\frac{d}{d\theta}$[tan($\theta$)]. $\frac{d}{d\theta}$[sec($\theta$)]=sec($\theta$)tan($\theta$) $\frac{d}{d\theta}$[tan($\theta$)]=$sec^{2}$($\theta$) The above derivatives could be derived using the quotient rule, but it'd be a good idea to have them memorized. We'll then find that f'($\theta$)=tan($\theta$)sec($\theta$)tan($\theta$)+sec($\theta$)$sec^{2}$($\theta$) We'll then factor out sec($\theta$) and get f'($\theta$)=sec($\theta$)($tan^{2}$($\theta$)+$sec^{2}$($\theta$))
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.