If f(x, y, z)=sin(3x-yz)?

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)?

df/dt=(df/dx)(dx/dt)+(df/dy)(dy/dt)+(df/dz)(dz/dt)

(3cos(3x-yz))(e^(t-1))+(cos(3x-z))(3t^2)

+(cos(3x-y))(1)

What should I do next to find df/dt(1)? Where do I plug in 1? Show your work through steps. Thanks.

Answer 1

Notice that df/dt is a function of one variable, t, so you will evaluate at t = 1.

In your expression df/dt = (3cos(3x-yz))(e^t-1)+(cos(3x-z))(3t²), you have x, y, and z which are given to you as functions of t. In other words, they are x(t), y(t), and z(t).

To make evaluation of df/dt at t=1 easier, you will want to evaluate x(1), y(1), and z(1). and plug them in for x, y, and z respectively.

See if that helps you.