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Math Help - [SOLVED] derivative help

  1. #1
    Member mybrohshi5's Avatar
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    [SOLVED] derivative help

    y=x^(cosx)

    i thought the derivative was....

    y'=ln(x)(x^cosx)(-sinx)

    where did i go wrong?

    thanks
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  2. #2
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     y=x^{\cos x}
    taking ln both side
     ln y= \cos x \ ln x

     \frac{d(ln y)}{dx}= \frac{d(\cos x\  ln x)}{dx}

     \frac{1}{y} \frac{dy}{dx}= \frac{1}{x} \cos x -ln x\  \sin x
     \frac{dy}{dx}= y\left ( \frac{1}{x} \cos x - ln x \ \sin x \right )
     \frac{dy}{dx}= x^{\cos x}\left ( \frac{1}{x} \cos x - ln x\ \sin x \right)
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  3. #3
    Member mybrohshi5's Avatar
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    never mind i got it. i was doing it way wrong lol.
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  4. #4
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    Quote Originally Posted by mybrohshi5 View Post
    y=x^(cosx)

    i thought the derivative was....

    y'=ln(x)(x^cosx)(-sinx)

    where did i go wrong?

    thanks
    y=x^{cos(x)}

    Take the ln of both sides.

    ln(y)=cos(x)ln(x)

    Use implicit differentiation:
    \frac{1}{y}\frac{dy}{dx}=-sin(x)ln(x)+cos(x)\frac{1}{x}

    \frac{dy}{dx}=-ysin(x)ln(x)+cos(x)\frac{1}{x}

    =-x^{cos(x)}sin(x)ln(x)+x^{cos(x)}cos(x)\frac{1}{x}
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  5. #5
    Member mybrohshi5's Avatar
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    Quote Originally Posted by ramiee2010 View Post
     y=x^{\cos x}
    taking ln both side
     ln y= \cos x ln x

     \frac{d(ln y)}{dx}= \frac{d(\cos x ln x)}{dx}

     \frac{1}{y} \frac{dy}{dx}= \frac{1}{x} \cos x -ln x \sin x
     \frac{dy}{dx}= y\left ( \frac{1}{x} \cos x - ln x \sin x \right )
     \frac{dy}{dx}= x^{\cos x}\left ( \frac{1}{x} \cos x - ln x \sin x \right)
    thats what i finally just did. i totally forgot about taking the natural log of both sides and differentiating haha. thanks for you help
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