Nicholas Hunter

2022-11-17

if $F\left(x,y\right)$ and $y=f\left(x\right)$,
$\frac{dy}{dx}=-\frac{\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(F\right)}{\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(F\right)}$
1) $F\left(x,y\right)$ 𝑎𝑛𝑑 $y=f\left(x\right)$ so his means that the function $F$ is a function of one variable which is $x$
2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constant

Neil Short

The function $F$ is by definition a function of two variables $x$ and $y$. It is trivial that we can take two partial derivatives of it.
This $F$ implicitly defines a function $f\left(x\right)$ of one variable by the constraint $F\left(x,y\right)=c$ where 𝑐 is a constant.
The correct equation for ${f}^{\prime }\left(x\right)$ you obtain from the chain rule:
$\frac{d}{dx}F\left(x,f\left(x\right)\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }x}F\left(x,f\left(x\right)\right)+\frac{\mathrm{\partial }}{\mathrm{\partial }y}F\left(x,f\left(x\right)\right){f}^{\prime }\left(x\right).$
Now observe that the LHS of this is zero because we said $F\left(x,f\left(x\right)\right)$ is constant. Can you solve this equation now for
${f}^{\prime }\left(x\right)=\frac{dy}{dx}\phantom{\rule{1em}{0ex}}?$

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