Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.2 The Derivative as a Function - Exercises - Page 115: 57

Answer

$$f'(x)=\frac{3}{2} x^{1 / 2}$$

Work Step by Step

Since $$ f^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ Then \begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{(x+h)^{3 / 2}-x^{3 / 2}}{h}\\ &=\lim _{h \rightarrow 0} \frac{\sqrt{(x+h)^{3}}-\sqrt{x^{3}}}{h} \frac{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}} \\ &=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h}\left(\frac{1}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}\right)\\ &= \lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h} \lim _{h \rightarrow 0}\left(\frac{1}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}\right)\\ &= \lim _{h \rightarrow 0} \frac{3x^2h+3xh^2+h^3}{h} \lim _{h \rightarrow 0}\left(\frac{1}{\sqrt{(x+h)^{3}}+\sqrt{x^{3}}}\right)\\ &= (3x^2)\frac{1}{2 \sqrt{x^{3}}}\\ &=\frac{3}{2} x^{1 / 2} \end{aligned}
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