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Calculating the derivative of a function of the form $e^{ax} \\ln(bx)$ using the product rule.

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Find the derivative of \\[ \\simplify{y=e^({a}x) ln({b}x)}. \\]

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If we have a function of the form $y=u(x)v(x)$, to calculate its derivative we need to use the product rule:

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\\[ \\dfrac{dy}{dx} = u(x) \\times \\dfrac{dv}{dx} + v(x) \\times\\dfrac{du}{dx}.\\]

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This can be split up into steps:

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    \n
  1. Identify the functions $u(x)$ and $v(x)$;
  2. \n
  3. Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
  4. \n
  5. Substitute these into the formula for the product rule to obtain an expression for $\\tfrac{dy}{dx}$;
  6. \n
  7. Simplify $\\tfrac{dy}{dx}$ where possible.
  8. \n
\n

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Following this process, we must first identify $u(x)$ and $v(x)$.

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As \\[ \\simplify{y=e^({a}x) ln({b}x)}, \\]

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let \\[ u(x) = \\simplify{e^({a}x)} \\quad \\text{and} \\quad v(x)=\\simplify{ln({b}x)}.\\]

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Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

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\\[ \\dfrac{du}{dx} = \\simplify{{a}e^({a}x)}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\dfrac{1}{x}.\\]

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Substituting these results into the product rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

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\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{du}{dx}\\times v(x) + u(x) \\times\\dfrac{dv}{dx} \\\\ \\\\&\\,=\\simplify{{a}e^({a}x)} \\times\\simplify{ln({b}x)} +\\simplify{e^({a}x)} \\times \\dfrac{1}{x}.  \\end{split}\\]

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Simplifying,

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\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\simplify{{a}e^({a}x) ln({b}x) + e^({a}x)/x} \\\\ &\\,= \\simplify{e^({a}x)({a}ln({b}x)+1/x)}.\\end{split} \\]

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$\\dfrac{dy}{dx}=$[[0]] 

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