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We use the PRODUCT RULE when the function that we need to differentiate is actually two functions multiplied together:

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If  $y=u \\times v$  then:

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\\[   \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} \\]

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Now it is time to get out your paper and pencil and try similar questions without the help. Carry out the same steps, lay it out the same way but you only need to input the final answer.

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We are asked to differentiate:

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\\[ y=\\var{aCF}x^{\\var{aP}} e^{\\var{eP}x} \\]

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Recognising that the function to differentiate is the product of two functions, we identify the two functions that are involved.

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$u$ is the first function, $v$ is second:

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$\\large u=\\var{aCF}x^{\\var{aP}} $                    $\\large v=e^{\\var{eP}x} $

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Now, we need to use the approriate techniques to differentiate each of these, for $u$ we can use the Power Rule and for $v$ your Table of Derivatives:

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This gives us:

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$\\large \\frac{du}{dx}=\\simplify{ {aP}*{aCF}*x^({aP}-1) }$          and          $ \\large \\frac{dv}{dx}= \\simplify{{eP}*e^({eP} x)}$

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We now use the formula:

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

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 Make the appropriate substitutions into the formula:

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$ \\large  \\frac{dy}{dx}= \\var{aCF}x^{\\var{aP}} \\times \\simplify{{eP}*e^({eP} x)} + e^{\\var{eP}x}  \\times \\simplify{{aP}*{aCF}*x^({aP}-1) } $

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Finally, we need to use our basic algebra to simplify this as much as possible. Multiply out any brackets where it would simplify and collect like terms:

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$  \\large \\frac{dy}{dx}=   \\simplify{{aCF}x^{{aP}} * {eP}*e^({eP} x)} +   \\simplify{{aP}*{aCF}*x^{{aP}-1} } e^{\\var{eP}x}  $

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Differentiate $  y=\\var{aCF}x^{\\var{aP}} e^{\\var{eP}x} $

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$  \\large \\frac{dy}{dx}=   $[[0]]

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