## Calculus 8th Edition

Published by Cengage

# Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 22

#### Answer

$D'(t)=\dfrac{-16t^2-3}{64t^4}$

#### Work Step by Step

$D(t)=\dfrac{1+16t^2}{(4t)^3}$ Use the quotient rule: $D'(t)=\dfrac{(64t^3)(0+32t)-(1+16t^2)(3)(4t)^2(4)}{(4t)^6}$ Apply exponents: $D'(t)=\dfrac{2048t^4-(1+16t^2)(3)(16t^2)(4)}{4096t^6}$ Perform multiplication: \begin{align} D'(t)&=\dfrac{2048t^4-(1+16t^2)(192t^2)}{4096t^6}\\&=\dfrac{2048t^4-192t^2-3072t^4}{4096t^6} \end{align} Combine like terms: $D'(t)=\dfrac{-1024t^4-192t^2}{4096t^6}$ Factor: $D'(t)=\dfrac{(64t^2)(-16t^2-3)}{4096t^6}$ Simplify: $D'(t)=\dfrac{-16t^2-3}{64t^4}$

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