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Math Help - Chain Rule-Partial Derivatives!

  1. #1
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    Unhappy Chain Rule-Partial Derivatives!

    Consider the function f(x,y)=x^2+y^2 where x=sin2θ and y=cos2θ
    df/dθ is given by?

    Hello

    Will someone be kind enough to provide me with the complete answer and explanation please?

    I have studied stuff like this in 8 years.I've running around this for the past three days.

    I have looked at the chain rule for multivariate functions.

    The Trig identity Sin^2+Cos^2=1

    I seem to be moving in circles
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pringles View Post
    Consider the function f(x,y)=x^2+y^2 where x=sin2θ and y=cos2θ
    df/dθ is given by?

    Hello

    Will someone be kind enough to provide me with the complete answer and explanation please?

    I have studied stuff like this in 8 years.I've running around this for the past three days.

    I have looked at the chain rule for multivariate functions.

    The Trig identity Sin^2+Cos^2=1

    I seem to be moving in circles
    By definition:

    \frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\cdot\frac{\,dx}{\,d\theta}+\frac{\partial f}{\partial y}\cdot\frac{\,dy}{\,d\theta}

    Since f\left(x,y\right)=x^2+y^2, when we partially differentiate the function wrt one variable, we hold the other variable constant. As a result, \frac{\partial f}{\partial x}=2x. I leave it for you to find \frac{\partial f}{\partial y}. You also need to find \frac{\,dx}{\,d\theta} and \frac{\,dy}{\,d\theta}. You can easily do these since x and y are defined in terms of theta, and also because this is ordinary differentiation.

    When you plug all those values into the equation, try to get your result in terms of one variable, \theta.

    Can you take it from here?
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  3. #3
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    Thumbs up I'll try

    Quote Originally Posted by Chris L T521 View Post
    By definition:

    \frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\cdot\frac{\,dx}{\,d\theta}+\frac{\partial f}{\partial y}\cdot\frac{\,dy}{\,d\theta}

    Since f\left(x,y\right)=x^2+y^2, when we partially differentiate the function wrt one variable, we hold the other variable constant. As a result, \frac{\partial f}{\partial x}=2x. I leave it for you to find \frac{\partial f}{\partial y}. You also need to find \frac{\,dx}{\,d\theta} and \frac{\,dy}{\,d\theta}. You can easily do these since x and y are defined in terms of theta, and also because this is ordinary differentiation.

    When you plug all those values into the equation, try to get your result in terms of one variable, \theta.

    Can you take it from here?
    I will give it a shot.Thanks
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