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Quotient rule

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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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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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\\(\\frac{df}{dx} = \\) [[0]]

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