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Math Help - chain rule

  1. #1
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    chain rule

    use the appropriate Chain rule to find the partial derivate of z with respect to u
    for z=(cos x) (sin y), x=u-v , y=u^2+v^2
    write your answer as a function of u and v.

    I dont understand how to write it in terms of u and v and i think i am messing up the partial derivatives can any one help me???
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by chris25 View Post
    use the appropriate Chain rule to find the partial derivate of z with respect to u
    for z=(cos x) (sin y), x=u-v , y=u^2+v^2
    write your answer as a function of u and v.

    I dont understand how to write it in terms of u and v and i think i am messing up the partial derivatives can any one help me???
    \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u} =-\sin(x)\sin(y)(1)+\cos(x)\cos(y)(2u)

    From here we know what x and y are as functions of u and v so just plug them in.

    \frac{\partial z}{\partial u}=\sin(u-v)\sin(u^2+v^2)(1)+\cos(u-v)\cos(u^2+v^2)(2u)

    Good luck.
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  3. #3
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    chain rule

    then you need to do the same with respect to v right?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jme44 View Post
    then you need to do the same with respect to v right?
    Yes, you can. Just apply the same concept.

    If f\equiv{f(x,y)}

    where

    x=x(u,v)

    and

    y=y(u,v)

    Then,

    \frac{\partial{f}}{\partial{v}}=\frac{\partial{f}}  {\partial{x}}\cdot\frac{\partial{x}}{\partial{v}}+  \frac{\partial{f}}{\partial{y}}\cdot\frac{\partial  {y}}{\partial{v}}
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