Finding the partial derivatives using the chain rule:

Quote:

Originally Posted by ^_^Engineer_Adam^_^

Find the indicated derivative by using the chain rule:
http://rogercortesi.com/eqn/tempimagedir/eqn2901.png

i think you mean you want du/dt and du/dz. All the d's here are partial derivatives, so you will write them as you see in the problem.

by the chain rule:

du/dt = du/dx*dx/dt + du/dy*dy/dt

=> du/dt = (ycos(xy))(2z*e^t) + (xcos(xy))(2t*e^-z)
=> du/dt = [t^2*e^-z(cos(2z*e^t*t^2*e^-z)][2z*e^t] + [2z*e^t(cos(2z*e^t*t^2*e^-z))][2t*e^-z]

...i plugged in the values for x and y, i'll leave you to simplify this, i have class soon

similarly, du/dz = du/dx*dx/dz + du/dy*dy/dz

=> du/dz = (ycos(xy))(2e^t) + (xcos(xy))(-t^2*e^-z)
=> du/dz = [t^2*e^-z(cos(2z*e^t*t^2*e^-z)][2e^t] + [2z*e^t(cos(2z*e^t*t^2*e^-z))][-t^2*e^-z]

...again, i leave you to simplify this