Let z=2x+y, x=t² and y=7t. At t=0, dx/dt=0. Sam says this makes dz/dt=0. Is Sam's claim true?

Circle one:   True   False

Assume x,y are differentiable at t and f is differentiable at (x(t),y(t)). For z=f(x(t),y(t)), where must the outer partials f_x and f_y be evaluated in the chain rule?

• At (dx/dt, dy/dt).
• At (x(t), dy/dt).
• At (x(t), y(t)).
• At (dx/dt, y(t)).

For differentiable z=f(x,y), x(t) and y(t), which general formula gives dz/dt? Here f_x=∂f/∂x and f_y=∂f/∂y.

• f_x(dx/dt) - f_y(dy/dt)
• f_x(dx/dt) + f_y(dy/dt)
• f_x(dy/dt) + f_y(dx/dt)

Let z = x squared + y squared with x = t and y = 2t. What is dz/dt at t = 1?

Answer: ______________

Let z=f(x(s,t),y(s,t)). At (s,t)=(a,b), x,y are differentiable; f is differentiable at P=(x(a,b),y(a,b)). At P, f_x=3 and f_y=-2. At (a,b), ∂x/∂s=4 and ∂z/∂s=18. Holding t fixed, find ∂y/∂s at (a,b).

• -3
• 3
• -15
• 15

Let z=x²+2y², x=s+t and y=st. What is ∂z/∂s at (s,t)=(1,3), with t held fixed?

Answer: ______________

At a time t0, x(t) and y(t) are differentiable, and f is differentiable at P=(x(t0),y(t0)). At P, f_x=-4 and f_y=3; at t0, dx/dt=2 and dy/dt=-5. For z=f(x(t),y(t)), what is dz/dt at t0?

Answer: ______________

For differentiable z=f(x,y), x(s,t) and y(s,t), which general formula gives ∂z/∂s while t is fixed? Here f_x=∂f/∂x and f_y=∂f/∂y.

• f_x(∂x/∂s) + f_y(∂y/∂s)
• f_x(∂x/∂t) + f_y(∂y/∂t)
• f_x(∂y/∂s) + f_y(∂x/∂s)

For differentiable z=f(x,y), x(s,t) and y(s,t), Leo claims ∂z/∂s=f_x ∂x/∂t+f_y ∂y/∂t is a valid general chain-rule formula. Is his claim true?

Circle one:   True   False

Let z = x squared + y with x = t and y = t squared. What is dz/dt at t = 3?

• 6
• 18
• 12