What is the implicit derivative of #1= x^2y-e^y-xy #?

Question

1501 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-01T17:38:20

The answer is #=-(2xy-y)/(x^2-e^y-x)#

Let

#f(x,y)=x^2y-e^y-xy-1#

Then,

The partial derivatives are

#(delf)/(delx)=2xy-y#

#(delf)/(dely)=x^2-e^y-x#

Therefore,

#dy/dx=-((delf)/(delx))/((delf)/(dely))#

#=-(2xy-y)/(x^2-e^y-x)#

You can calculate by direct differentiation

#2xy+x^2dy/dx-e^ydy/dx-xdy/dx-y=0#

#dy/dx(x^2-e^y-x)=-2xy+y#

#dy/dx=-(2xy-y)/(x^2-e^y-x)#