Question

已知z=z(x,y)由sin(xy)+z^2=sin(x+z)确定,求z关于x的偏导数

Answer 1

sin(xy)+z^2=sin(x+z)

ycos(xy)+2z (δz/δx) = [ 1+(δz/δx) ] cos(x+z)

[2z - cos(x+z)](δz/δx) = cos(x+z) - ycos(xy)

δz/δx = [cos(x+z) - ycos(xy)]/[2z - cos(x+z)]