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Thread: Chain rule with several variables

  1. #1
    MHF Contributor arbolis's Avatar
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    Chain rule with several variables

    Calculate $\displaystyle \frac{\partial z}{\partial x}$ using the chain rule where $\displaystyle z=f(u,v,r)$, $\displaystyle u=g(x,y,r)$, $\displaystyle v=h(x,y,r)$ and $\displaystyle r=k(x,y)$.
    My attempt : $\displaystyle \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}+\frac{\partial z}{\partial r}\cdot \frac{\partial r}{\partial x}$. Is that enough? Or have I to calculate $\displaystyle \frac{\partial u}{\partial x}$ and other terms?
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    Quote Originally Posted by arbolis View Post
    Calculate $\displaystyle \frac{\partial z}{\partial x}$ using the chain rule where $\displaystyle z=f(u,v,r)$, $\displaystyle u=g(x,y,r)$, $\displaystyle v=h(x,y,r)$ and $\displaystyle r=k(x,y)$.
    My attempt : $\displaystyle \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}+\frac{\partial z}{\partial r}\cdot \frac{\partial r}{\partial x}$. Is that enough? Or have I to calculate $\displaystyle \frac{\partial u}{\partial x}$ and other terms?
    Yes --- because

    $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial g}{\partial x}+\frac{\partial g}{\partial r} \frac{\partial r}{\partial x}$
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