## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises: 14

#### Answer

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

#### Work Step by Step

The product rule states that if f(z)=h(z)g(z), then f'(z)=h(z)g'(z)+g(z)h'(z). Since f(z)=ztan(z), we can set h(z)=z and g(z)=tan(z), then apply the product rule to find f'(z). Therefore, f'(z)=(z)$\frac{d}{dz}$[tan(z)]+(tan(z))$\frac{d}{dz}$[z] We know that $\frac{d}{dz}$[z]=1, using the power rule $\frac{d}{dz}$[tan(z)]=$sec^{2}$(z), which can be derived by using the quotient rule to compute the derivative of the equation g(z)=$\frac{sin(z)}{cos(z)}$, which of course is equivalent to tan(z) Therefore, f'(z)=(z)($sec^{2}$(z))+(tan(z))(1) =z$sec^{2}$(z)+tan(z)

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