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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}\$$

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\$$\\frac{df}{dx} = \$$ [[0]]

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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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Quotient rule

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