## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises: 31

#### Answer

$$f'(x)=4x^3$$ The domain of both $f(x)$ and $f'(x)$ is $R$

#### Work Step by Step

$$f(x)=x^4$$ According to definition, $$f'(x)=\lim\limits_{h\to0}\frac{(x+h)^4-x^4}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{[(x+h)^2-x^2][(x+h)^2+x^2]}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{(x+h-x)(x+h+x)[(x+h)^2+x^2]}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{h(x+h+x)[(x+h)^2+x^2]}{h}$$ $$f'(x)=\lim\limits_{h\to0}(2x+h)[(x+h)^2+x^2]$$ $$f'(x)=2x[x^2+x^2]$$ $$f'(x)=4x^3$$ The domain of both $f(x)$ and $f'(x)$ is $R$

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