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

df/dt = (df/dx)*(dx/dt)+(df/dy)*(dy/dt)+(df/dz)*(dz/dt) by the chain rule. Once you have taken the six derivatives, you plug in one for t.

Which of the six derivatives are you having trouble taking, or was the concept the problem?

## Comments

How do you find df/dx and dx/dt?

df/dx =d(sin(3x-c))/dx. We know that sin(x) = (e^ix - e^-ix)/2i. Can you solve from there?

x = e^(t-1) = (e^t)/e, dx/dt = (e^t)/e same as before.