# If z = f(x, y), where f is differentiable, and the following applies, find dz/dt when t = 1.?

x = g(t) y = h(t)

g(1) = -5 h(1) = 1

g’ (1) = -6 h’ (1) = 1

fx(-5, 1) = 1 fy(-5, 1) = -1

no thats not the right answer

• laughing is correct up til

"

dz/dt = -6 1 + 1 -1

dz/dt = -6 – 2

dz/dt = -8

"

It’s easy to fix though (just 1 * -1 = -1, not -2):

dz/dt = -6 1 + 1 -1

dz/dt = -6 – 1

dz/dt = -7

• If z = f(x,y)

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

Where dz/dx and dz/dy are partial derivatives.

I am not sure what fx(-5,1) = 1 and fy(-5,1) = -1 mean, but I guess they represent these partial derivatives. Summarizing the information until now:

g(1) = -5 –> x = -5 in this special case for t = 1

h(1) = 1 –> y = 1 in this special case for t = 1

g'(1) = -6

h'(1) = 1

for those specific values of x and y, the partial derivatives are:

dz/dx = 1

dz/dy = -1

Filling in gives:

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

dz/dt = -6 1 + 1 -1

dz/dt = -6 – 2

dz/dt = -8

Edit: I have to admit, that’s a bit stupid :). Thanks for correcting me!