I have a function
u(r,t)= (sin(kr-wt))/r
How do I find the partial derivatives:
du/dr
d^2u/dr^2
d^2u/dt^2
Thanks in advance
$\displaystyle u(r,t)= (sin(kr-wt))/r$
for $\displaystyle \frac{du}{dr}$
treat "t" as a constant, and then find derivative.
so
we find:
$\displaystyle \frac{du}{dt}(r^{-1}sin(kr-wt))= -r^{-2}sin(kr-wt)+r.r^{-1}cos(kr-wt)$
$\displaystyle = cos(kr-wt)- \frac{sin(kr-wt)}{r^{2}}$
Now from there solve the rest...