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Math Help - log deriv.

  1. #1
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    log deriv.

    Use logarithmic differentiation to find the derivative of the function.
    y = x^sinx

    How do you do this? Help please.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kwivo View Post
    Use logarithmic differentiation to find the derivative of the function.
    y = x^sinx

    How do you do this? Help please.
    begin by taking the log of both sides.

    y = x^{\sin x}

    \Rightarrow \ln y = \ln x^{\sin x}

    \Rightarrow \ln y = \sin x \ln x

    differentiating implicitly with repsect to x, we get:

    \frac {y'}y = \cos x \ln x + \frac {\sin x}x

    \Rightarrow y' = y \left( \cos x \ln x + \frac {\sin x}x\right) = x^{\sin x}\left( \cos x \ln x + \frac {\sin x}x\right)
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  3. #3
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    Hello, kwivo!

    Use logarithmic differentiation to differentiate: . y \:=\:x^{\sin x}

    We have: . y \;=\;x^{\sin x}

    Take logs: . \ln y \:=\:\ln\left(x^{\sin x}\right)\quad\Rightarrow\quad\ln y \:=\:\sin x\cdot\ln x

    Differentiate implicitly: . \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\sin x\cdot\frac{1}{x} + \cos x\cdot\ln x

    Multiply by y\!:\;\;\frac{dy}{dx}\;=\;y\left[\frac{\sin x}{x} + \cos x\!\cdot\!\ln x\right]

    Replace y\!:\;\;\frac{dy}{dx}\;=\;x^{\sin x}\left[\frac{\sin x}{x} + \cos x\!\cdot\!\ln x\right]

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  4. #4
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    Quote Originally Posted by kwivo View Post
    Use logarithmic differentiation to find the derivative of the function.
    y = x^sinx
    I'll use the never well weighed trick a=e^{\ln a},\,\forall a>0, then y=e^{\sin x\cdot\ln x}

    Now it remains to compute (\sin x\cdot\ln x)', and yields the desired.
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