Posted by Brian on .
If f(x, y, z)=sin(3xyz), where x=e^(t1), y=t^3, z=t2, what's df/dt(1)?
x(1)=1 y(1)=1 z(1)=1
df/dt=(3cos(3xyz))(e^(t1))+(cos(3xz))(3t^2)+(cos(3xy))(1)
I seriously need help on this and tell me the answer and show your work through steps. Thanks.

Urgent math 
Steve,
using x' for dx/dt, etc,
dF/dt = Fx x' + Fy y' + Fz z'
= 3cos(3xyz)(e^(t1))  zcos(3xyz)(3t^2)  ycos(3xyz)(1)
at t=1,
dF/dt = 3cos(4)(1) + cos(4)(3)  cos(4)(1)
= 5cos(4)
Hard to see what the difficulty is; they gave you the formula for df/dt, as well as the values for t,x,y,z. Just plug them in. If you don't see how to come up with df/dt, review partial derivatives some more.