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Quotient rule

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\\(\\frac{df}{dx} = \\) [[0]]

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

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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}x^{\\var{b}}+\\var{f}\\)   and   \\(v=\\var{c}cos(\\var{d}x)\\)

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then   \\(\\frac{du}{dx}=\\var{b}*\\var{a}x^{\\var{b}-1}\\)   and   \\(\\frac{dv}{dx}=-\\var{d}*\\var{c}sin(\\var{d}x)\\)

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Putting these results together as shown in the rule gives:

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

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\\(\\frac{df}{dx}=\\frac{\\simplify{({c}*cos({d}x))*{b}*{a}x^{{b}-1}+({a}x^{{b}}+{f})*({c}*{d}*sin({d}x))}}{(\\var{c}*cos(\\var{d}x))^2}\\)

", "statement": "

Differentiate the function:

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

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