Confusing use of the chain rule

Question

Let $f$ be a function of three real variables, $(x, y, z)$ where $z$ is a function of $(x,y)$. By the chain rule on $f(x,y,z(x,y))$:

$$\frac{\partial{f}}{\partial{x}}=\frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{x}}+\frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{x}}+\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{x}}=\frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{x}}$$

But this implies $\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{x}}=0$! Which of course can't be true. Something in my understanding is missing. Maybe these are different "kinds" of derivatives?

What is the solution to resolve the confusion?

Answer 1

You problem will become clearer if you note : $$\frac{\partial{f\big(x,y,z(x,y)\big)}}{\partial{x}}=\frac{\partial{f}}{\partial{x'}}\big(x,y,z(x,y)\big)\frac{\partial{x}}{\partial{x}}+\frac{\partial{f}}{\partial{y'}}\big(x,y,z(x,y)\big)\frac{\partial{y}}{\partial{x}}+\frac{\partial{f}}{\partial{z'}}\big(x,y,z(x,y)\big)\frac{\partial{z}}{\partial{x}}.$$ Indeed, you are not differentiating among the same variables : writting $f\big(x,y,z(x,y)\big)=f\circ g(x,y)$ with $g(x,y)=\big(x,y,z(x,y)\big),$ and so $(x,y)\overset{g}{\mapsto}\big(x,y,z(x,y)\big)\overset{f}{\mapsto} f\big(x,y,z(x,y)\big),$ then $x$ is the first variable in the source of $g$ (i.e. $\mathbb{R}^2$) while $x'$ is the first variable in the source of $f$ (i.e. $\mathbb{R}^3$).

Answer 2

Be careful with the notations $$\frac{df(x,y,z)}{dx}=f_x(x,y,z)=f_x\ne\frac{d}{dx}\{f(x,y,z(x,y))\}=\\f_x+f_zz_x=f_x(x,y,z(x,y))+f_z(x,y,z(x,y))\frac{dz(x,y)}{dx}$$