## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 13

#### Answer

f'(x)=sec(x)((2$x^{4}$-4$x^{-1}$)(tan(x)+(8$x^{3}$+4$x^{-2}$))

#### 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). Since f(x)=(2$x^{4}$-4$x^{-1}$)sec(x), we can set h(x)=(2$x^{4}$-4$x^{-1}$) and g(x)=sec(x), then apply the product rule to find f'(x). Therefore, f'(x)=(2$x^{4}$-4$x^{-1}$)$\frac{d}{dx}$[sec(x)]+(sec(x))$\frac{d}{dx}$[2$x^{4}$-4$x^{-1}$] We know that $\frac{d}{dx}$[2$x^{4}$-4$x^{-1}$]=8$x^{3}$+4$x^{-2}$, using the power rule $\frac{d}{dx}$[sec(x)]=sec(x)tan(x), which can be derived by using the quotient rule to compute the derivative of the equation g(x)=$\frac{1}{cos(x)}$, which of course is equivalent to sec(x) Therefore, f'(x)=(2$x^{4}$-4$x^{-1}$)(sec(x)tan(x))+(sec(x))(8$x^{3}$+4$x^{-2}$) =sec(x)((2$x^{4}$-4$x^{-1}$)(tan(x)+(8$x^{3}$+4$x^{-2}$))

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