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Math Help - Logramethic differentaion

  1. #1
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    Logramethic differentaion

    Can you differentiate this please? I keep getting the wrong answers.

    Determine the question of the tangent line to y = x^sinx AT x = pi/2

    This seemed alot easier when looking at it
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    Quote Originally Posted by proski117 View Post
    Can you differentiate this please? I keep getting the wrong answers.

    Determine the question of the tangent line to y = x^sinx AT x = pi/2

    This seemed alot easier when looking at it
     \ln(y) = \ln(x^{\sin(x)})

     \ln(y) = \sin(x) \ln(x)

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

     \frac{dy}{dx} \times \frac{d}{dy} \ln(y) = \frac{d}{dx} \sin(x) \ln(x)

     \frac{dy}{dx} \times \frac{1}{y} = \frac{d}{dx} \sin(x) \ln(x)

    Now use the product rule on the RHS, substitude the original equation in for 'y', and rearrange to get  \frac{dy}{dx} = .
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  3. #3
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    Quote Originally Posted by proski117 View Post
    Can you differentiate this please? I keep getting the wrong answers.

    Determine the question of the tangent line to y = x^sinx AT x = pi/2

    This seemed alot easier when looking at it
    y = x^{\sin x} \Rightarrow \ln y = \sin x \ln x.

    The derivative of the left hand side is \frac{1}{y} \cdot \frac{dy}{dx}. Use the product rule to differentiate the right hand side.

    Substitute x = \frac{\pi}{2} and y = \frac{\pi}{2} (why?) and solve for the value of m = \frac{dy}{dx}.

    You now have the gradient of the tangent and you know a point on the tangent. Getting the equation of the tangent should be routine.
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  4. #4
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    Quote Originally Posted by proski117 View Post
    Can you differentiate this please? I keep getting the wrong answers.

    Determine the question of the tangent line to y = x^sinx AT x = pi/2

    This seemed alot easier when looking at it
    logarithmic derivative ...

    y = x^{\sin{x}}

    \ln{y} = \ln\left(x^{\sin{x}}\right)

    \ln{y} = \sin{x} \cdot \ln{x}

    take the derivative of both sides ...

    \frac{y'}{y} = \sin{x} \cdot \frac{1}{x} + \cos{x} \cdot \ln{x}

    y' = y\left(\sin{x} \cdot \frac{1}{x} + \cos{x} \cdot \ln{x}\right)

    y' = x^{\sin{x}}\left(\sin{x} \cdot \frac{1}{x} + \cos{x} \cdot \ln{x}\right)<br />

    sub in \frac{\pi}{2} for x and evaluate.
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  5. #5
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    The question said do not simplify my answer...but when you guys showed me the derivative, i subbed in pi/2 and got 1 as my answer for slope. Did I sub in wrong?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by proski117 View Post
    The question said do not simplify my answer...but when you guys showed me the derivative, i subbed in pi/2 and got 1 as my answer for slope. Did I sub in wrong?
    yes, the slope is 1 at that point
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