## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.1 Exercises - Page 103: 24

#### Answer

$f'(x)=-\dfrac{2}{x\sqrt{x}}$

#### Work Step by Step

$f(x)=\dfrac{4}{\sqrt{x}}$ The derivative of $f$ at $x$ is $f'(x)=\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$ Find $f(x+\Delta x)$ by substituting $x$ by $x+\Delta x$ in $f(x)$: $f(x+\Delta x)=\dfrac{4}{\sqrt{x+\Delta x}}$ Substitute $f(x+\Delta x)$ and $f(x)$ into the formula and evaluate: $f'(x)=\lim_{\Delta x\to0}\dfrac{\dfrac{4}{\sqrt{x+\Delta x}}-\dfrac{4}{\sqrt{x}}}{\Delta x}=...$ $...=\lim_{\Delta x\to0}\dfrac{\dfrac{4\sqrt{x}-4\sqrt{x+\Delta x}}{\sqrt{x^{2}+x\Delta x}}}{\Delta x}=...$ $...=\lim_{\Delta x\to0}\dfrac{4\sqrt{x}-4\sqrt{x+\Delta x}}{\Delta x\sqrt{x^{2}+x\Delta x}}=...$ Rationalize the numerator: $...=\lim_{\Delta x\to0}\dfrac{4(\sqrt{x}-\sqrt{x+\Delta x})}{\Delta x\sqrt{x^{2}+x\Delta x}}\cdot\dfrac{\sqrt{x}+\sqrt{x+\Delta x}}{\sqrt{x}+\sqrt{x+\Delta x}}=...$ $...=\lim_{\Delta x\to0}\dfrac{4[(\sqrt{x})^{2}-(\sqrt{x+\Delta x})^{2}]}{\Delta x(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$ $...=\lim_{\Delta x\to0}\dfrac{4(x-x-\Delta x)}{\Delta x(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$ $...=\lim_{\Delta x\to0}\dfrac{-4\Delta x}{\Delta x(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$ $...=\lim_{\Delta x\to0}\dfrac{-4}{(\sqrt{x^{2}+x\Delta x})(\sqrt{x}+\sqrt{x+\Delta x})}=...$ $...=\dfrac{-4}{[\sqrt{x^{2}+x(0)}][\sqrt{x}+\sqrt{x+0}]}=\dfrac{-4}{(\sqrt{x^{2}})(\sqrt{x}+\sqrt{x})}=...$ $...=\dfrac{-4}{(x)(2\sqrt{x})}=-\dfrac{2}{x\sqrt{x}}$ $f'(x)=-\dfrac{2}{x\sqrt{x}}$

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