1=x/y-e^(xy)1=xy−exy
First we have to know that we can differentiate each part separately
Take y=2x+3y=2x+3 we can differentiate 2x2x and 33 seperately
dy/dx=dy/dx2x+dy/dx3 rArrdy/dx=2+0dydx=dydx2x+dydx3⇒dydx=2+0
So similarly we can differentiate 11, x/yxy and e^(xy)exy separately
dy/dx1=dy/dxx/y-dy/dxe^(xy)dydx1=dydxxy−dydxexy
Rule 1: dy/dxC rArr 0dydxC⇒0 derivative of a constant is 0
0=dy/dxx/y-dy/dxe^(xy)0=dydxxy−dydxexy
dy/dxx/ydydxxy we have to differentiate this using the quotient rule
Rule 2: dy/dxu/v rArr ((du)/dxv-(dv)/dxu)/v^2dydxuv⇒dudxv−dvdxuv2 or (vu'-uv')/v^2
u=x rArr u'=1
Rule 2: y^n rArr (ny^(n-1)dy/dx)
v=y rArr v'=dy/dx
(vu'+uv')/v^2=(1y-dy/dxx)/y^2
0=(1y-dy/dxx)/y^2-dy/dxe^(xy)
Lastly we have to differentiate e^(xy) using a mixture of the chain and the product rule
Rule 3: e^u rArr u'e^u
So in this case u=xy which is a product
Rule 4: dy/dxxy=y'x+x'y
x rArr 1
y rArr dy/dx
y'x+x'y=dy/dxx+y
u'e^u=(dy/dxx+y)e^(xy)
0=(1y-dy/dxx)/y^2-(dy/dxx+y)e^(xy)
Expand out
0=(1y-dy/dxx)/y^2-dy/dxxe^(xy)+ye^(xy)
Times both sides by y^2
0=y-dy/dxx-dy/dxxe^(xy)y^2+ye^(xy)y^2
0=y-dy/dxx-dy/dxxe^(xy)y^2+e^(xy)y^3
Place all the dy/dx terms on one side
y-e^(xy)y^3=dy/dxx-dy/dxxe^(xy)y^2
Factorize out dy/dx on the RHS (right hand side)
-y-e^(xy)y^3=dy/dx(x-xe^(xy)y^2)
(-(y+e^(xy)y^3))/(x-xe^(xy)y^2)=dy/dx
