Question

If f(x, y, z)=sin(3x-yz), where x=e^(t-1), y=t^3, z=t-2, what's df/dt(1)?

x(1)=1 y(1)=1 z(1)=-1
df/dt=(3cos(3x-yz))(e^(t-1))+(cos(3x-z))(3t^2)+(cos(3x-y))(1)

I seriously need help on this and tell me the answer and show your work through steps. Thanks.

Answer 1

using x' for dx/dt, etc,

dF/dt = Fx x' + Fy y' + Fz z'
= 3cos(3x-yz)(e^(t-1)) - zcos(3x-yz)(3t^2) - ycos(3x-yz)(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.