Partial Derivative and chain rule...

**Raidan** · May 3rd 2009, 04:44 AM

Hi guys I am having difficulty while revising..

Consider the function r=(x^2+y^2+z^2) and f=r^-1

Find Partial derivatives rx,ry,rz. I have done this part but having difficulty doing this part

Use chain rule to find fx,fy,fz and hence show that grad ∆f =(-r^-3)(r) where r=xi+yj+zk

**Scott H** · May 3rd 2009, 06:33 AM

Here, the Chain Rule states

$\frac{\partial f}{\partial x}=\frac{df}{dr}\frac{\partial r}{\partial x}.$

We have, by calculation

$\frac{df}{dr} = -\frac{1}{r^2}\;\;\;\;\;\;\;\;\;\;\frac{\partial r}{\partial x} = 2x.<br />$

Thus, we have

$\frac{\partial f}{\partial x}=\left(-\frac{1}{r^2}\right)\cdot(2x)=-\frac{2x}{r^2}.$

I'm not sure what you mean by $\mbox{grad}\,\Delta f$ , but for $\nabla f$ I calculated something that looked quite different from the answer given in the problem.

**Raidan** · May 3rd 2009, 06:37 AM

Originally Posted by Scott H

Here, the Chain Rule states

$\frac{\partial f}{\partial x}=\frac{df}{dr}\frac{\partial r}{\partial x}.$

We have, by calculation

$\frac{df}{dr} = -\frac{1}{r^2}\;\;\;\;\;\;\;\;\;\;\frac{\partial r}{\partial x} = 2x.<br />$

Thus, we have

$\frac{\partial f}{\partial x}=\left(-\frac{1}{r^2}\right)\cdot(2x)=-\frac{2x}{r^2}.$

I'm not sure what you mean by $\mbox{grad}\,\Delta f$ , but for $\nabla f$ I calculated something that looked quite different from the answer given in the problem.

I see thanx...actually grad f is same as saying $\nabla f$ I didnt know how to use the symbol...

**Raidan** · May 3rd 2009, 06:49 AM

oo the function is actually supposed to be r=(x^2+y^2+z^2)^1/2 that is probably why you didnt get the same answer as the one I posted...sorry I missed that..and thanx

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