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Math Help - find dy/dx by implicit differentiation

  1. #1
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    find dy/dx by implicit differentiation

    Can someone show me step by step how to do

    y= sin xy
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  2. #2
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    Quote Originally Posted by Nightasylum View Post
    Can someone show me step by step how to do

    y= sin xy
    y = \sin(xy)

    y' = \cos(xy) \cdot (xy' + y)

    y' = xy'\cos(xy) + y\cos(xy)

    y' - xy'\cos(xy) = y\cos(xy)

    y'[1 - x\cos(xy)] = y\cos(xy)

    y' = \frac{y\cos(xy)}{1 - x\cos(xy)}
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  3. #3
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    Quote Originally Posted by Nightasylum View Post
    Can someone show me step by step how to do

    y= sin xy
    first we need to use the chain rule on the right hand side

    \frac{dy}{dx}=\cos(xy)\cdot \frac{d}{dx}(xy)

    Now we need to use the product rule to finish taking the derivative

    \frac{dy}{dx}=\cos(xy)\cdot \left( 1 \cdot y+x\frac{dy}{dx} \right)=y\cos(xy)+x\cos(xy)\frac{dy}{dx}

    Now we just need to solve for the derivative

    \frac{dy}{dx}-x\cos(xy)\frac{dy}{dx}=y\cos(xy)

    \frac{dy}{dx}(1-x\cos(xy))=y\cos(xy)

    \frac{dy}{dx}=\frac{y\cos(xy)}{1-x\cos(xy)}
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  4. #4
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    Im still not grasping how to do this , I see you get the derivative of sin times the stuff in the ( ) right? but where does the +Y come from
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  5. #5
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    Quote Originally Posted by Nightasylum View Post
    Im still not grasping how to do this , I see you get the derivative of sin times the stuff in the ( ) right? but where does the +Y come from
    it's the product rule ...

    the derivative of (xy) w/r to x is ...

    x \cdot y' + y\cdot 1
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  6. #6
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    Quote Originally Posted by Nightasylum View Post
    Im still not grasping how to do this , I see you get the derivative of sin times the stuff in the ( ) right? but where does the +Y come from

    The implict derivative of

    \frac{d}{dx}(xy)

    uses both the product rule and chain rule

    First the product rule give

    \frac{d}{dx}(dx)=\left( \frac{d}{dx}x\right) \cdot y + x\left( \frac{d}{dx} y\right)

    i.e the dervitive of the first times the 2nd plus the derviatve of the 2nd times the first

    Now using implict differenation we get and

    Remembering that \frac{d}{dx}y=\frac{dy}{dx}=y'

    \frac{d}{dx}(dx)=\left( \frac{d}{dx}x\right) \cdot y + x\left( \frac{d}{dx} y\right)=1\cdot y+x\cdot \frac{dy}{dx}=y+xy'
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  7. #7
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    Thanks guys I see what im doing now wish I had some tutor here that was as helpful as you all. Last time I made a stop by the tutor on campus they had no clue what I was doing and my professor only allows 30 mins during their office hours because she trys to help everyone. /sigh
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