# Thread: Chain rule with several variables

1. ## 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?

2. Originally Posted by arbolis
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

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