i'm dealing with a really simple question that I'm struggling to find an answer because I'm messed up with physics notations. Suppose you have the relation $x=Ky$ and you have a function $f$. $K$ is a fixed scalar. You know explicitly $f$ as well as $$ f'\left(x\right)=\frac{\partial f}{\partial x}\left(x\right)=\frac{\partial f}{\partial x}\left(Ky\right) $$ I just want to find $$ \frac{\partial f }{\partial y}\left(x\right) $$ in terms of $x$, is it right to write $$ \frac{\partial f }{\partial y}\left(x\right)=K\frac{\partial f}{\partial x}\left(x\right) \ \ \ ? $$
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$\frac{\partial f }{\partial y}\left(x\right)$ uses partial derivatives, and seems to ask what is the rate of change of $f(x)$ with respect to $y$ if $x$ is held constant, despite knowing $y=Kx$. – Henry Oct 22 '21 at 17:05
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What you are having is (in clearer notation) $f(g(y))$, where $f(x)$ is 'some' function, and $g(y)=Ky$. So you only have to apply the chain rule $$ \frac{d}{dy}f(g(y)) = f'(x)\Big|_{x=y} g'(y) = K f'(x)\Big|_{x=y} $$ Note that you don't need partial derivatives, because everything is a function of a single variable. Note also that $f'(x)\Big|_{x=y} $ is often abbreviated into $f'(y)$.... but perhaps this way it's clearer.
Pierpaolo Vivo
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It is the derivative of $f(x)$ with respect to its argument, evaluated at $x = y$. For example, if $f(x)=x^2-x$, then $f'(x) = 2x-1$, and $f'(x)\Big|_{x=y}=2y-1$ – Pierpaolo Vivo Oct 23 '21 at 14:16
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