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We are asked to differentiate:
\n\\[ y=\\var{aCF}x^{\\var{aP}} e^{\\var{eP}x} \\]
\n\nRecognising that the function to differentiate is the product of two functions, we identify the two functions that are involved.
\n\n$u$ is the first function, $v$ is second:
\n\n$\\large u=\\var{aCF}x^{\\var{aP}} $ $\\large v=e^{\\var{eP}x} $
\n\n
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:
\n\nThis gives us:
\n$\\large \\frac{du}{dx}=\\simplify{ {aP}*{aCF}*x^({aP}-1) }$ and $ \\large \\frac{dv}{dx}= \\simplify{{eP}*e^({eP} x)}$
\n\nWe now use the formula:
\n$ \\large \\frac{dy}{dx}=u \\frac{dv}{dx} + v \\frac{du}{dx} $
\nMake the appropriate substitutions into the formula:
\n\n$ \\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) } $
\n\n\n
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:
\n\n$ \\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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First identify the two functions $u$ and $v$:
\n$u=$[[0]] $v=$[[1]]
\n\n
Now differentiate each one:
\n$ \\large \\frac{du}{dx}= $[[2]] $ \\large\\frac{dv}{dx}= $[[3]]
\n\n
Substitute each component into the formula in the correct place:
\n$ \\large \\frac{dy}{dx}=$[[4]]$ \\large \\times$[[5]]$ \\large + $[[6]]$ \\large \\times$[[7]]
\n\n
Finally tidy this up to give your final answer:
\n$ \\large \\frac{dy}{dx}= $[[8]]
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