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Thread: 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
    $\displaystyle \ln(y) = \ln(x^{\sin(x)}) $

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

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

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

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

    Now use the product rule on the RHS, substitude the original equation in for 'y', and rearrange to get $\displaystyle \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
    $\displaystyle y = x^{\sin x} \Rightarrow \ln y = \sin x \ln x$.

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

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

    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 ...

    $\displaystyle y = x^{\sin{x}}$

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

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

    take the derivative of both sides ...

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

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

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

    sub in $\displaystyle \frac{\pi}{2}$ 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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    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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