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So I stumbled into this problem, and I honestly have no idea what rules to use. I'd appreciate some help, and if you could direct me to some useful resource to learn the right method I'd be thankful.

Here's the question :

$y = f(x,p) \quad ; x = g(p,y)$

Calculate the derivative of y with respect to x in terms of the 4 partial derivatives we suppose known :

$\dfrac{\partial f}{\partial x} \quad ; \quad \dfrac{\partial f}{\partial p} \quad ; \quad \dfrac{\partial g}{\partial p} \quad ; \quad \dfrac{\partial g}{\partial y}$

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$$y = f(x,p) \quad ;\quad dy=\dfrac{\partial f}{\partial x} dx+ \dfrac{\partial f}{\partial p}dp$$

$$x = g(p,y) \quad ;\quad dx=\dfrac{\partial g}{\partial p} dp+ \dfrac{\partial g}{\partial y}dy$$

$$dp=\frac{dy-\dfrac{\partial f}{\partial x} dx}{\dfrac{\partial f}{\partial p}} = \frac{ dx- \dfrac{\partial g}{\partial y}dy}{\dfrac{\partial g}{\partial p}}$$

$$\dfrac{\partial g}{\partial p}\left(dy-\dfrac{\partial f}{\partial x} dx\right) = \dfrac{\partial f}{\partial p}\left(dx-\frac{ \partial g}{\partial y}dy\right)$$

$$\dfrac{\partial g}{\partial p}\left(\frac{dy}{dx}-\dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial p}\left(1-\frac{ \partial g}{\partial y}\frac{dy}{dx}\right)$$

$$\frac{dy}{dx}\left(\dfrac{\partial g}{\partial p} + \dfrac{\partial f}{\partial p}\frac{ \partial g}{\partial y} \right)= \dfrac{\partial g}{\partial p}\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial p} $$

$$\frac{dy}{dx} = \frac{\dfrac{\partial g}{\partial p}\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial p}}{\dfrac{\partial g}{\partial p} + \dfrac{\partial f}{\partial p}\dfrac{ \partial g}{\partial y} }$$

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  • $\begingroup$ Ok. I see you did not need to use any specific rule, just simple computations ... Thanks a lot, I appreciate it. $\endgroup$ – Cerberos26 Mar 17 at 15:52

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