- #1
brentd49
- 74
- 0
So, the question is: find the gradient of f=ln(r), where r=sqrt(x^2+y^2+z^2)^(1/2).
For the partial with respect to x, I use the chain rule: df/dr*dr/dx.
df/dr=1/r
dr/dx=(1/2)*(2x)=x
Which would give df/dx = x/r.
But the book gets x/r^2
Where does the extra 1/r come from?
For the partial with respect to x, I use the chain rule: df/dr*dr/dx.
df/dr=1/r
dr/dx=(1/2)*(2x)=x
Which would give df/dx = x/r.
But the book gets x/r^2
Where does the extra 1/r come from?