First we have to familrise ourselves with some calculs rules
f(x)=2x+4 we can differentiate 2x and 4 separately
f'(x)=dy/dx2x+dy/dx4=2+0=2
Similarly we can differentiate the 4, y and -(x-e^y)/(y-x) separately
dy/dx4=dy/dxy-dy/dx(x-e^y)/(y-x)
We know that differentiating constants dy/dx4=0
0=dy/dxy-dy/dx(x-e^y)/(y-x)
Likewise the rule for differentiating y is dy/dxy=dy/dx
0=dy/dx-dy/dx(x-e^y)/(y-x)
Lastly to differentiate (x-e^y)/(y-x) we have to use the quotient rule
Let x-e^y=u
and
Let y-x=v
The quotient rule is (vu'-uv')/v^2
(du)/dx=(du)/dxx-(du)/dxe^y
When deriving e we use the chain rule such that e^y rArr (du)/dxe^y
so u'=1-dy/dxe^y
y-x=v
so
v'=(dv)/dxy-(dv)/dxx
Using the same rules from above it becomes
v'=dy/dx-1
Now we have to do the quotient rule
(vu'-uv')/v^2=((y-x)(1-(dy)/dxe^y)-(x-e^y)(dy/dx-1))/(y-x)^2
0=dy/dx-((y-x)(1-(dy)/dxe^y)-(x-e^y)(dy/dx-1))/(y-x)^2
Expand out
0=dy/dx-((y-ydy/dxe^y-x+xdy/dxe^y)-(xdy/dx-x-e^ydy/dx+e^y))/(y-x)^2
0=dy/dx-(y-ydy/dxe^y-x+xdy/dxe^y-xdy/dx+x+e^ydy/dx-e^y)/(y-x)^2
Multiply both sides by (y-x)^2
0=dy/dx(y-x)^2-(y-ydy/dxe^y+xdy/dxe^y-xdy/dx+e^ydy/dx-e^y)
0=dy/dx(y-x)^2-y+ydy/dxe^y-xdy/dxe^y+xdy/dx-e^ydy/dx+e^y
Place all the dy/dx terms on one side
y-e^y=dy/dx(y-x)^2+ydy/dxe^y-xdy/dxe^y+xdy/dx-e^ydy/dx
Factories dy/dx out of every term
y-e^y=dy/dx((y-x)^2+ye^y-xe^y+x-e^y)
(y-e^y)/((y-x)^2+ye^y-xe^y+x-e^y)=dy/dx
f'(x)=(y-e^y)/((y-x)^2+ye^y-xe^y+x-e^y)
