• Jan 23rd 2010, 10:37 AM
ur5pointos2slo
Find gradient f at the given point.

f(x,y,z)= ( (x^2+y^2+z^2)^(-1/2) ) + ln(xyz) , (-1,2,-2)

df/dx = -x/(x^2+y^2+z^2)^(3/2)
df/dy = -y/(x^2+y^2+z^2)^(3/2)
df/dz = -z/(x^2+y^2+z^2)^(3/2)

ok so after finding the partials with respect to x,y,z and plugging in the point I get the answer

-1/27 i - 2/27 j + 2/27 k

-26/27 i +23/54 j -23/54 k
Can anyone verify which answer is correct?
• Jan 23rd 2010, 10:56 AM
drumist
Did you forget to take the partial derivatives of $\displaystyle \ln(xyz)$?
• Jan 23rd 2010, 11:00 AM
ur5pointos2slo
Quote:

Originally Posted by drumist
Did you forget to take the partial derivatives of $\displaystyle \ln(xyz)$?

I thought the partial derivative of ln(xyz) is 0 with respect to all variables?
• Jan 23rd 2010, 11:02 AM
drumist
Nope.

$\displaystyle \frac{\partial}{\partial x} \ln(xyz) = \frac{yz}{xyz} = \frac{1}{x}$
• Jan 23rd 2010, 11:04 AM
ur5pointos2slo
Quote:

Originally Posted by drumist
Nope.

$\displaystyle \frac{\partial}{\partial x} \ln(xyz) = \frac{yz}{xyz} = \frac{1}{x}$

O yes. Thanks