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Product rule

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Differentiate the following using the product rule:

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

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\$$\\frac{dy}{dx}=\$$[[0]]

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

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Recall the product rule if \$$y=u \\times v\$$ where \$$u\$$ and \$$v\$$ are both functions of \$$x\$$ then

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\$$\\frac{dy}{dx}=v \\times \\frac{du}{dx}+u \\times \\frac{dv}{dx}\$$

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

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

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Substituting into \$$\\frac{dy}{dx}=v \\times \\frac{du}{dx}+u \\times \\frac{dv}{dx}\$$ gives

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

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Which tidies up to \$$\\frac{dy}{dx}= \\var{a}\\var{g}x^\\var{a-1}(\\var{c}x^{\\var{d}} + \\var{b}) + \\var{g}x^\\var{a} (\\var{c}\\var{d}x^{\\var{d-1}} + \\var{b})\$$

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