Understanding transition from full derivative to partial

1. Oct 21, 2015

I was looking over a derivation to find the laplacian from cartesian to cylindrical and spherical coordinates here: http://skisickness.com/2009/11/20/

Everything seems fine, but there is an instance (I have attached a screenshot) where implicit differentiation is done to find $$\frac {\partial \phi}{\partial x}$$ by finding $$\frac {d\phi}{dx}$$ implicitly and then equating this to $$\frac {\partial \phi}{\partial x}$$ It seems like they are two separate equations. In this instance, x was independent of y, but in a future scenario, if x and y were dependent (e.g. y = kx), then the implicit differentiation would give a different result, right? In that case, would this approach work to still give $$\frac {d\phi}{dx} = \frac {\partial \phi}{\partial x} {\text ?}$$

I understand the strategy based on the above equation (in the link) where they are trying to find each respective partial derivative to go from cartesian to cylindrical coordinates, but equating the full derivative to the partial derivative seems a little off. It looks like it works in this particular case because x and y are independent, but if they were dependent, would this approach work?

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2. Oct 21, 2015

Geofleur

It might help to look at a really simple example, like $f(x,y)=xy$. Then implicit differentiation wrt $x$ gives $df = ydx$. But this change in $f$ assumes $y$ is held constant, so a better way to write it might be $(df)_y = y (dx)$. Then "dividing through by $dx$" just yields $\frac{\partial f}{\partial x} = y$. If $y$ doesn't depend on $x$ then $\frac{dy}{dx}=0$, so that $\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=\frac{\partial f}{\partial x}=y$.

An important thing to keep in mind is that partial derivatives don't "care" about functional relationships that may exist between the arguments. Let's take the previous example but suppose that $y$ depends on $x$ this time as, say, $y = x^3$. We would still have $\frac{\partial f}{\partial x} = y$. On the other hand, we would now write $\frac{df}{dx} = \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx} = y + 3x^3 = 4x^3.$ Because $\frac{\partial f}{\partial x} = y = x^3$, we no longer have $\frac{\partial f}{\partial x}=\frac{df}{dx}$.

Last edited: Oct 21, 2015