## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 127: 56

#### Answer

If $g$ is differentiable at $t=7$ then $3g$ is also differentiable at $t=7$.

#### Work Step by Step

$g(t)$ is differentiable at $t=7$ means that $$\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$$ has a defined value. Now about $3g$, the derivative of $3g$, $(3g)'$, at $t=7$ has the following formula: $$(3g)'=\lim_{h\to0}\frac{(3g)(h+7)-(3g)(7)}{h}=\lim_{h\to0}\frac{3\Big(g(h+7)\Big)-3(g(7))}{h}$$ $$(3g)'=\lim_{h\to0}3\Big(\frac{g(h+7)-g(7)}{h}\Big)=3\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$$ Since $\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$, as said above, has a defined value, $(3g)'=3\lim_{h\to0}\frac{g(h+7)-g(7)}{h}$ also has a defined value. This means $3g$ is also differentiable at $t=7$.

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.