Testing the multivariable chain rule

Question

Let $x = t$, $y = t^2$, and $z = xy$. Then $$\frac{\partial y}{\partial t} = 2t$$ and since $t = x$, $$\frac{\partial y}{\partial x} = 2x = 2t.$$ Everything is consistent. But now consider using the chain rule \begin{align}\frac{\partial y}{\partial x} &= \frac{\partial\left[z(x,y)/x\right]}{\partial x}\\ &= \frac{\partial\left(z/x\right)}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial\left(z/x\right)}{\partial z}\frac{\partial z}{\partial x} \\ &= -\frac{z}{x^2}\cdot1+\frac{1}{x}\cdot y \\ &= -\frac{xy}{x^2}+\frac{y}{x} \\ &= 0 \end{align} Certainly, then, something has gone very wrong. What is my conceptual error here?

Answer 1

The problem is that $y$ does not depend explicitly of $x$. You have

$x=x(t)=t$, $y=y(t,x)=t^2$, and $z=z(x,y)=xy$. Then

$$ \frac{\partial y}{\partial t}=2t $$ $$ \frac{\partial y}{\partial x}=0 $$ Note that $$ \frac{d}{dt}y(t,x)=\frac{\partial y}{\partial t}+ \frac{\partial y}{\partial x}\frac{\partial x}{\partial t} =\frac{\partial y}{\partial t} $$ Finally $$ \frac{\partial y}{\partial x}=\frac{\partial\left[z(x,y)/x\right]}{\partial x}=\frac{y}{x}-z\frac{1}{x^2}=0 $$

Answer 2

Actually $y$ being dependent on $t,$ depends on $x.$ But in the second passage (by using the chain rule) variable $y$ is assumed as independent on $x,$ so the result is predictable.