1. ## Vector Differentiation etc.

Given that V = xyz(r^−7), find the first partial derivative of V (wrt x) and deduce that the second partial derivative of V (wrt to x) is

−21xyz(r^−9) + (63x^3)yz(r^−11)

NB: d = delta sign, del = del function (upside down triangle)

I've tried it several times now. I can see where the coefficents will eventually come from being multiples of 7, but I can't get the answer above. Can anyone help? Thanks.

2. Originally Posted by Yerobi
Given that V = xyz(r^−7), find the first partial derivative of V (wrt x) and deduce that the second partial derivative of V (wrt to x) is

−21xyz(r^−9) + (63x^3)yz(r^−11)

NB: d = delta sign, del = del function (upside down triangle)

I've tried it several times now. I can see where the coefficents will eventually come from being multiples of 7, but I can't get the answer above. Can anyone help? Thanks.
What is r? Is it $\displaystyle r = \sqrt{x^2 + y^2 + z^2}$?

-Dan

3. Indeed.

4. a preliminary calculation:

$\displaystyle r = \sqrt {x^2 + y^2 + z^2 } \Rightarrow \frac{{\partial r}} {{\partial x}} = \frac{x} {r}$

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$\displaystyle \begin{gathered} V = xyzr^{ - 7} \hfill \\ \hfill \\ \frac{{\partial V}} {{\partial x}} = yzr^{ - 7} - 7xyzr^{ - 8} \frac{{\partial r}} {{\partial x}} = yzr^{ - 7} - 7x^2 yzr^{ - 9} \hfill \\ \end{gathered}$

thus:

$\displaystyle \frac{{\partial ^2 V}} {{\partial x^2 }} = - 7yzr^{ - 8} \frac{{\partial r}} {{\partial x}} - 14xyzr^{ - 9} + 63x^2 yzr^{ - 10} \frac{{\partial r}} {{\partial x}} = - 21xyzr^{ - 9} + 63x^3 yzr^{ - 11}$

5. Thanks! The next part of the Q is:

Use symmetry to write down the corresponding expressions for the second derivatives of V wrt to y and z.

Deduce that (del)^2(V) = 0.

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The second derivative of V wrt y would be the same but using dr/dx = y/r etc. yeah? Once I have derivatives wrt y and z, would their sum equal 0?