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Math Help - Derivative of Trignometric Equation

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Derivative of Trignometric Equation

    Find y' if sin(2xy)+cos^2(xy)=x^2y^2+3y^2.

    I tried using Implicit Differentiation, but the answer I got was all complicated & I can't simplify it.
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    Quote Originally Posted by StarlitxSunshine View Post
    Find y' if sin(2xy)+cos^2(xy)=x^2y^2+3y^2.

    I tried using Implicit Differentiation, but the answer I got was all complicated & I can't simplify it.
    The derivatives of the terms on the left side are as follows:

    1st term: cos(2xy)\frac{d}{dx}(2xy)=(2y+2xy')cos(2xy)

    2nd term: \frac{d}{dx}cos^2(xy)=2cos(xy)\frac{d}{dx}cos(xy)

    See if you can find the derivative \frac{d}{dx}cos(xy).

    Is everything I've done so far consistent with what you've attempted?
    Last edited by adkinsjr; November 8th 2009 at 02:16 PM.
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  3. #3
    Junior Member StarlitxSunshine's Avatar
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    Actually, I didn't find the derivative of the first term of the first term.

    Sin (2xy) I simply put as Cos (2xy). Why are you taking the derivative of the inside function ?
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  4. #4
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    Quote Originally Posted by StarlitxSunshine View Post
    Actually, I didn't find the derivative of the first term of the first term.

    Sin (2xy) I simply put as Cos (2xy). Why are you taking the derivative of the inside function ?
    Chain Rule
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    Quote Originally Posted by StarlitxSunshine View Post
    Actually, I didn't find the derivative of the first term of the first term.

    Sin (2xy) I simply put as Cos (2xy). Why are you taking the derivative of the inside function ?
    Chain rule, just as e^i*pi said. Always remember to use the chain rule whenever you see a function in the parenthesis. If it isn't just x, then you have to apply the chain rule. In this case, you have to apply a combination of the chain rule and implicit differentiation. So try to find the derivative \frac{d}{dx}cos(xy)
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  6. #6
    Junior Member StarlitxSunshine's Avatar
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    Oh. Whenever it isn't x...I think I get it.

    Okay, that should be:

    <br />
\frac{d}{dx}[cos(xy)] = -sin(xy)((x)(y')+(y)(1))<br />

    <br />
= -xy'sin(xy) -ysin(xy)<br />

    ...Right ?
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    Quote Originally Posted by StarlitxSunshine View Post
    Oh. Whenever it isn't x...I think I get it.

    Okay, that should be:

    <br />
\frac{d}{dx}[cos(xy)] = -sin(xy)((x)(y')+(y)(1))<br />

    <br />
= -xy'sin(xy) -ysin(xy)<br />

    ...Right ?
    Yup
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  8. #8
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    So do you understand how I found the derviative of cos^2(xy)

    ???

    That one is kind of confusing. Other than that, the right side of the equation is very basic. I'm sure you can figure that out. So now you just have to solve for y'.
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  9. #9
    Junior Member StarlitxSunshine's Avatar
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    Yup Yup, Sorry :33

    I understood that part & was able to finish the question myself.

    Thank you very much =)
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