Help me find the implicit differentiation of this equation

(x^2)y-x(e^x)+2=0


I don't understand implicit differentiation at all, so please explain each step to me!

(x^2)y-x(e^x)+2=0

Take the differential of each term:
d(x²y) - d(xe^x) + d(2) = d(0)

Apply product rule of differentiation and the fact that the differential of a constant is zero:
yd(x²) + x²dy - xd(e^x) - e^xdx + 0 = 0
y(2xdx) + x²dy - xe^xdx - e^xdx = 0
2xydx + x²dy - e^x(x+1)dx = 0

Define y' = dy/dx and divide by dx:
2yx + x²y' - e^x(x+1) = 0

If you need the expression for y' then solve the equation for y':
y' = [e^x(x+1) - 2xy]/x²

(x^2)y-x(e^x)+2=0

First term:
(x^2)y

We differentiate with respect to x using the multiplication rule:

2xy + x^2(dy/dx)

Second term:

-x(e^x)

-(e^x + xe^x)

Third term is constant so the derivative is zero. The right side is 0 so the derivative is 0.

2xy + x^2(dy/dx) -(e^x + xe^x) = 0

Now solve for dy/dx

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

(x^2)y' + 2xy - x*e^(x) - e^(x) = 0

(x^2)y' = [e^(x) + x*e^(x) - 2xy]

y' = [e^(x) + x*e^(x) - 2xy]/x^2

x^2 dy/dx + 2xy - xe^x - e^x = 0
x^2 dy/dx = e^x + xe^x - 2xy
dy/dx = (e^x + xe^x - 2xy)/x^2

2x*y dx - x^2 dy - e^x dx - xe^x dx=0
dx(2xy -e^x - xe^x) + dy(x^2)=0
dy(x^2)=dx(e^x+xe^x-2xy)
dy/dx=[e^x+xe^x-2xy]/x^2