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Differentiate the function:

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\$$f(x)=\\frac{\\var{a} + \\var{g}x^{\\var{b}}}{\\var{c}x^{\\var{d}} + \\var{f}}\$$

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Quotient rule

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\$$\\frac{df}{dx} = \$$ [[0]]

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\$$f(x)=\\frac{\\var{a} +\\var{g}x^{\\var{b}}}{\\var{c}x^{\\var{d}} +\\var{f}}\$$

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Recall the quotient rule: if \$$y=\\frac{u}{v}\$$ where \$$u\$$ and \$$v\$$ are both functions of \$$x\$$ then

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\$$\\frac{dy}{dx}=\\frac{v\\frac{du}{dx}-u\\frac{dv}{dx}}{v^2}\$$

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Let   \$$u=\\var{a} +\\var{g}x^{\\var{b}}\$$   and   \$$v=\\var{c}x^{\\var{d}} +\\var{f}\$$

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then   \$$\\frac{du}{dx}=\\var{g}\\var{b}x^{\\var{b-1}}\$$   and   \$$\\frac{dv}{dx}=\\var{d}\\var{c}x^{\\var{d-1}}\$$

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Putting these results together as shown in the rule gives:

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\$$\\frac{df}{dx}=\\frac{(\\var{c}x^{\\var{d}} +\\var{f}) \\times \\var{g}\\var{b}x^{\\var{b-1}} - (\\var{a} +\\var{g}x^{\\var{b}}) \\times \\var{d}\\var{c}x^{\\var{d-1}}}{(\\var{c}x^{\\var{d}} +\\var{f})^2}\$$

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\$$\\frac{df}{dx}=\\frac{\\var{g}\\var{b}x^{\\var{b-1}}(\\var{c}x^{\\var{d}} +\\var{f}) - \\var{d}\\var{c}x^{\\var{d-1}}(\\var{a} +\\var{g}x^{\\var{b}}) }{(\\var{c}x^{\\var{d}} +\\var{f})^2}\$$

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