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Differentiate the following using the product rule:

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

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Product rule

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\\(\\frac{dy}{dx}=\\)[[0]]

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

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

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

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

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