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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 $0$?

Nick
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1 Answers1

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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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W.R.P.S
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  • view $g(x,y)=f(x,y,u(x,y))$, then what should be $\frac{\partial g}{\partial x}$? I am confused with these operations – Nick Dec 03 '16 at 16:37
  • @Nick Liu ${∂g\over∂x}$ that is PARTIAL DERIVATIVES z=f(x,y) the partial derivative with respect to x is just the ordinary derivative of the function of a single variable that we get by keeping y fixed, but chain rule is result of differentiable function .if we have a z=F(u,v) u=$\varphi$(x,y) , v=$\phi$(x,y) then we have ${∂z\over∂x}$=${∂f\over∂u}$${∂u\over∂x}$+${∂f\over∂v}$${∂v\over∂x}$ and similarly for y – W.R.P.S Dec 03 '16 at 17:25
  • but in this form we have two function with x,y variable and we can change x when y is fixed and then our function is just on x,but in your question we can't do change in x when other variable take fixed – W.R.P.S Dec 03 '16 at 17:25
  • i'm sorry for my english language mistakes .i hope it can help you – W.R.P.S Dec 03 '16 at 17:30