Glenn Hopkins

2022-07-16

Find the derivative of the functions by using the Quotient Rule. Simplify your answer.
1.$f\left(x\right)=\frac{x+1}{x-1}$
2. $f\left(x\right)=\frac{{x}^{2}-3x+4}{x+2}$
3. $\frac{\sqrt{x}-6}{\sqrt{x}+6}$

phravincegrln2

1. $f\left(x\right)=\frac{x+1}{x-1}$
Using the Quotient rule
${f}^{\prime }\left(x\right)=\left(x-1\right)\frac{d}{dx}\frac{\left(x+1\right)-\left(x+1\right)\frac{d}{dx}\left(x-1\right)}{\left(x-1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(x-1\right)-x-1}{\left(x-1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{-2}{\left(x-1{\right)}^{2}}$
2. $f\left(x\right)=\frac{{x}^{2}-3x+4}{x+2}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{\left(x+2\right)\frac{d}{dx}\left({x}^{2}-3x+4\right)-\left({x}^{2}-3x+4\right)\frac{d}{dx}\left(x+2\right)}{\left(x+2{\right)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{\left(x+2\right)\left(2x-3\right)-\left({x}^{2}-3x+4\right)}{\left(x+2{\right)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{2{x}^{2}-3x+4x-6-{x}^{2}+3x-4}{\left(x+2{\right)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{{x}^{2}+4x-10}{\left(x+2{\right)}^{2}}$
3. $\frac{\sqrt{x}-6}{\sqrt{x}+6}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left[\frac{\sqrt{x}-6}{\sqrt{x}+6}\right]=\frac{\left(\sqrt{x}+6\right)\frac{d}{dx}\left(\sqrt{x}-6\right)-\left(\sqrt{x}-6\right)\frac{d}{dx}\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+6{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\left(\sqrt{x}+6\right)\frac{1}{\sqrt[2]{x}}-\left(\sqrt{x}-6\right)\frac{1}{\sqrt[2]{x}}}{\left(\sqrt{x}+6{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{6}{\sqrt{x}\left(\sqrt{x}+6{\right)}^{2}}$

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