Partial Differentiation, complication in variables held constant

maple
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Hi,

this is a pretty trivial question. would be grateful if someone could answer it for me.

in spherical polars x=rcos(theta)sin(PHI) and so on for y, and z

Now, why is

d/dr= dx/dr*d/dx + dy/dr*d/dy+ dz/dr*d/dz

where everything is partial. dx/dr, dy/dr and dz/dr at partial derivates held at contant thetha and phi.

why are they held at constant thetha and phi?

r^2=x^2 + y^2 + z^2

so r=function of (x,y,z)

thus we can write this out as an exact differential we get:

dr=dr/dx*dr + dr/dy*dy + dr/dz*dz

dr/dx is held constant wrt y,z etc. and NOT thetha and phi.

Can some explain how the differential at the top works.

thanks
 
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"constant theta & phi"...It's nothing special,really.It's the plain old CHAIN RULE for partial derivatives.An implicit dependence of variables requires this chain rule...U should write

f\left(x(r,\vartheta,\varphi),y(r,\vartheta,\varphi),z(r,\vartheta)\right)

and the simply apply the chain rule...

Partial differentiation is ordinary differentiation,but with some of the variables "kept constant" while evaluating that limit implied by the def.

Daniel.
 
thanks daniel,

can you actually spell all this out? I'm not sure how to put what you said into practice.

Many thanks.
 
"f" depends on r implicitely,by means of the functions x(r,...),y(r,...),z(r,...).So applying the chain rule,one gets simply

\frac{\partial f}{\partial r} =\frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial r}

and the same for the other 2 implicit variables.

U can use the diffeomorphism

\left \{ \begin{array}{c} x=r\sin\vartheta \cos\varphi \\ y=r\sin\vartheta \sin\varphi \\ z=r\cos\vartheta

together with

f\left(x\left(r,\vartheta,\varphi\right),y\left(r,\vartheta,\varphi\right),z\left(r,\vartheta\right)\right) = x^{2}+z^{3}y+\sqrt{xyz}

to find the 3 partial derivatives wrt the implicit variables...


Daniel.
 
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