Question

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

Answer 1

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.