How do you implicitly differentiate #11=(x-y)(e^y+e^x)#?

1 Answer
Apr 2, 2016

Use the power rule and some algebra to get #dy/dx=-(e^y+e^x+xe^x-ye^x)/(xe^y-e^y-e^x-ye^y)#.

Explanation:

Begin by taking the derivative with respect to #x# on both sides:
#d/dx(11)=d/dx((x-y)(e^y+e^x))#

The left side is easy - the derivative of a constant is #0#:
#0=d/dx((x-y)(e^y+e^x))#

The right side is a little harder. Because there's multiplication going on there, we'll have to apply the product rule, which says:
#d/dx(uv)=u'v+uv'#

In our case,
#u=x-y->u'=1-dy/dx#
#v=e^y+e^x->v'=e^ydy/dx+e^x#

Applying the power rule:
#d/dx((x-y)(e^y+e^x))=(x-y)'(e^y+e^x)+(x-y)(e^y+e^x)'#
#color(white)(XX)=(1-dy/dx)(e^y+e^x)+(x-y)(e^ydy/dx+e^x)#

Doing some algebra to isolate #dy/dx#:
#0=(1-dy/dx)(e^y+e^x)+(x-y)(e^ydy/dx+e^x)#
#color(white)(XX)=e^y+e^x-e^ydy/dx-e^xdy/dx+xe^ydy/dx+xe^x-ye^ydy/dx-ye^x#
#color(white)(XX)=(xe^ydy/dx-e^ydy/dx-e^xdy/dx-ye^ydy/dx)+(e^y+e^x+xe^x-ye^x)#
#-(e^y+e^x+xe^x-ye^x)=xe^ydy/dx-e^ydy/dx-e^xdy/dx-ye^ydy/dx#
#-(e^y+e^x+xe^x-ye^x)=dy/dx(xe^y-e^y-e^x-ye^y)#
#-(e^y+e^x+xe^x-ye^x)/(xe^y-e^y-e^x-ye^y)=dy/dx#

And there you have it.