## Calculus 8th Edition

Published by Cengage

# Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 140: 11

#### Answer

$F'(r)=\frac{-15}{r^{4}}$

#### Work Step by Step

$F(r)=\frac{5}{r^{3}}$ Lets use the formula $(\frac{f}{g})'=\frac{gf'-fg'}{g^{2}}$ then $F'(r)=(\frac{5}{r^{3}})'=\frac{r^{3} \times (5)' - 5 \times (r^{3})'}{r^{6}}$ by the formulas $(x^{n})'=nx^{n-1}$, $(c)' = 0$ $(r^{3})' = 3r^{2}$ $(5)' = 0$ therefore $F(r)=\frac{5}{r^{3}}=\frac{r^{3} \times (5)' - 5 \times (r^{3})'}{r^{6}}=\frac{0- 5 \times 3r^{2}}{r^{6}}=\frac{-15r^{2}}{r^{6}}=\frac{-15}{r^{4}}$

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