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Thread: Logarithmic Differentiation

  1. #1
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    Logarithmic Differentiation

    Find f'(x) using logarithmic differentiation:
    f(x)=(4x+4)^(3x)
    Thanks for the help.
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  2. #2
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    Hello, nystudent2729!

    Find $\displaystyle f'(x)$ using logarithmic differentiation: .$\displaystyle f(x)\:=\:(4x+4)^{3x}$
    We have: .$\displaystyle y \;=\;(4x+4)^{3x}$

    Take logs: .$\displaystyle \ln(y) \;=\;\ln(4x+4)^{3x}$

    . . . . . . . . $\displaystyle \ln(y) \;=\;3x\!\cdot\!\ln(4x+4)$


    Differentiate: .$\displaystyle \frac{1}{y}\!\cdot\!\frac{dy}{dx} \;=\;3x\!\cdot\!\frac{4}{4x+4} + 3\!\cdot\!\ln(4x+4) \;=\;3\bigg[\frac{x}{x+1} + \ln(4x+4)\bigg] $

    . . Then: .$\displaystyle \frac{dy}{dx} \;=\;3y\bigg[\frac{x}{x+1} + \ln(4x+4)\bigg] $


    Therefore: .$\displaystyle \frac{dy}{dx} \;=\;3(4x+4)^{3x}\left[\frac{x}{x+1} + \ln(4x+4)\right] $

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