Partial derivative chain rule problem

Question

Suppose $f(x,y,z)$ is a function, where $z$ is a function of $x,y$, i.e. $z=u(x,y)$. Does the following hold due to the chain rule?

$$\frac{\partial }{\partial x}f(x,y,u(x,y))=\frac{\partial }{\partial x}f(x,y,u(x,y))+\frac{\partial }{\partial u}f(x,y,u(x,y))\frac{\partial u}{\partial x}$$

I dont think this is true because if so then $\frac{\partial }{\partial u}f(x,y,u(x,y))\frac{\partial u}{\partial x}$ should be $0$?

Answer 1

you can't get ∂ from your function beacuse with change in x then z change too, so you should use defferential

${d\over dx}$f(x,y,u(x,y))=${∂\over∂x}$f(x,y,u(x,y))+${∂\over∂u}$f(x,y,u(x,y))${∂u\over∂x}$

chain rule definition chain rule

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You problem will become clearer if you note :

${∂f(x,y,z(x,y))\over∂x}$=${∂f\over∂x′}$(x,y,z(x,y))${∂x\over∂x}$+${∂f\over∂y′}$ (x,y,z(x,y))${∂y\over∂x}$+${∂f\over∂z′}$(x,y,z(x,y))${∂z\over∂x}$.

Indeed, you are not differentiating among the same variables : writting f(x,y,z(x,y))=f∘g(x,y) with g(x,y)=(x,y,z(x,y)),and so (x,y)↦(x,y,z(x,y))↦f(x,y,z(x,y)),then x is the first variable in the source of g (i.e. ℝ2) while x′ is the first variable in the source of f (i.e. ℝ3).

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