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If we have $y(x,t)=x^3+t^3$ and $x=t^2$ then $$\frac{\partial y}{\partial s} = \frac{\partial y}{\partial t}\frac{\partial t}{\partial s}+\frac{\partial y}{\partial x}\frac{\partial x}{\partial s}$$ Then setting $s=t$ we get $$\frac{\partial y}{\partial t}=\frac{\partial y}{\partial t}+\frac{\partial y}{\partial x}\frac{\partial x}{\partial t}$$ and we see that $$\frac{\partial y}{\partial x}\frac{\partial x}{\partial t}=0$$ which doesn't seem to be correct. I am confused what actually happens here?

Kimari
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Well in your second equality the y on the left in partial y with respect to t refers to the composite function $Y(t) = y(t^2,t)$ while the y to the right of the equality refers to the original function $y(x,t) =x^3 +t^3$ so you can't cancel them .The s is redundant and never should have entered the problem .

user439545
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  • I have heard that when we do a partial derivative, we simply neglect any other variables. So, I wonder how you know that we have to change $y=x^3+t^3$ to $y=t^6+t^3$ on the left before doing the differentiation? – Kimari Nov 14 '17 at 11:21
  • The composite function must get a different name from the original function .also writing y=y(x,t) is standard practice but is nevertheless an abuse of language that usually causes no difficulty to the initiated but is not strictly speaking correct . If you had written G(x,t) =$x^3+t^3$in the first place then Y and G are clearly different . – user439545 Nov 19 '17 at 04:21
  • Also when you write partial of y with respect to t you really mean partial of either Y or G with respect to t and you cant tell from the print which is ,if you see y in both cases :the user has to know which it is from experience ;the advantage of this kind of language abuse (or slang) is the saving of letters ,more important to the thinking then you might think. – user439545 Nov 19 '17 at 04:29