Question #8d476

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Answer 1 · 2016-03-02T19:35:01

#f_x = ze^(xz)+y#

#f_y = x#

When we differentiate with respect to a given variable, all other variables are treated as constants. Thus, given
#f(x, y, z) = e^(xz)+xy#

#" "#

#f_x = d/dx(e^(xz)+xy)#

#=(d/dxe^(xz))+(d/dx(xy))#

#=ze^(xz)+y#

and

#f_y = d/dy(e^(xz)+xy)#

#=(d/dye^(xz))+(d/dyxy)#

#=x#