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Consider three quantities related by: $$y=f(t), x=g(t)$$ Then what is $(\frac{\partial y}{\partial x})_t$? Is it simply that we cannot perform such an operation or is it $0$ or $\infty$? If it is one of the latter cases then what is: $(\frac{\partial x}{\partial y})_t$ and is this equal to $[(\frac{\partial y}{\partial x})_t]^{-1}$?

Quantum spaghettification
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1 Answers1

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You could use the chain rule:

$(\frac{\partial y}{\partial x})=(\frac{\partial y}{\partial t})(\frac{\partial t}{\partial x})={f'(t)\over g'(t)}$

and it is equal to $(\frac{\partial x}{\partial y})^{-1}$

cr001
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  • In this how are you holding $t$ constant? – Quantum spaghettification Nov 27 '15 at 06:50
  • $t$ is not a constant and both $x$ and $y$ are depending on $t$ based on your description in the question. Am I not understanding correctly? – cr001 Nov 27 '15 at 06:55
  • You are correct, however, when I perform the partial derivative I want to keep $t$ constant, so kind of like taking the partial derivative of a constant w.r.t a constant. – Quantum spaghettification Nov 27 '15 at 06:59
  • When you fix $t$, $f'(t)\over g'(t)$ becomes a constant as well then your partial derivative is just a straight line, is this what you meant? – cr001 Nov 27 '15 at 07:01
  • I think I understood what you are asking now, the problem here is that $y$ is not a function of both $x$ and $t$, you could write $t=g^{-1}(x)$ and write $y=f(g^{-1}(x))$ and now there is no $t$ left and you will see what is happening. – cr001 Nov 27 '15 at 07:03
  • No not really, let us say you have $y=y(x,t)$ I want to find $(\frac{\partial y}{\partial x})_t$ but where $y$ has no $x$ dependency. You would nataully think that $(\frac{\partial y}{\partial x})_t=0$ but that rises the question of what is $(\frac{\partial x}{\partial y})_t$ – Quantum spaghettification Nov 27 '15 at 07:05
  • The problem is that you don't have $y(x,t)$, you only have $y(x)$ or $y(t)$ as $x$ and $t$ are not free variables. – cr001 Nov 27 '15 at 07:07
  • If you meant $(\frac{\partial y}{\partial x})\over{\partial t} $ that is a function of $t$ as well which is not necessarily $0$ and $(\frac{\partial x}{\partial y})\over{\partial t} $ is another function of $t$ and the product of the two is not one. – cr001 Nov 27 '15 at 07:11