Question

How do you implicitly differentiate `-1=(x+y)^2-xy-e^(3x+7y)`?

Answer 1

`y'={2x+y-3e^(3x+4y)}/(4e^(3x+4y)-x-2y).`

Explanation:

Given Eqn. is `-1=(x+y)^2-xy-e^(3x+7y),` or,
`xy+e^(3x+4y)=(x+y)^2+1.`

Diff. both sides,

`(xy)'+{e^(3x+\color{red}{7}y)}'={(x+y)^2}'+0.`
`:. xy'+yx'+e^(3x+7y)(3x+7y)'=2(x+y)(x+y)'`
`:.xy'+y+e^(3x+7y){(3x)'+(7y)'}=2(x+y)(x'+y')`
`:. xy'+y+e^(3x+7y)(3+7y')=2(x+y)(1+y'),` i.e.,
`xy'+y+3e^(3x+4y)+7y'e^(3x+7y)=2(x+y)+2y'(x+y).`
`:. xy'+7y'e^(3x+7y)-2y'(x+y)=2(x+y)-y-3e^(3x+4y).`
`:. y'(x+7e^(3x+7y)-2x-2y)=2x+2y-y-3e^(3x+4y),` or,
`y'(7e^(3x+7y)-x-2y)=2x+y-3e^(3x+4y).`

Hence, `y'={2x+y-3e^(3x+7y)}/(7e^(3x+7y)-x-2y).`

`y'` can further be simplified, as below :-
For this, we write the given eqn. as `e^(3x+7y)=(x+y)^2+1-xy=x^2+xy+y^2+1,` & submit the value of e^(3x+7y)) in `y'` to give,

`y'={2x+y-3(x^2+xy+y^2+1)}/{4(x^2+xy+y^2+1)-x-2y}.`

Answer 2

`y'={2x+y-3(x^2+y^2+xy+1)}/{7(x^2+y^2+xy+1)-(2y+x)}.`

Explanation:

We will use the given eqn. as `e^(3x+7y)=(x+y)^2-xy+1=x^2+y^2+xy+1,` & diff. its both sides, to get,

`e^(3x+7y)*(3+7y')=2x+2yy'+xy'+y.`
Subbing `x^2+y^2+xy+1` for `e^(3x+7y)` in `L.H.S.,`

`3(x^2+y^2+xy+1)+7y'(x^2+y^2+xy+1)=2x+y+(2y+x)y'`

`y'{7(x^2+y^2+xy+1)-(2y+x)}=2x+y-3(x^2+y^2+xy+1)`

`y'={2x+y-3(x^2+y^2+xy+1)}/{7(x^2+y^2+xy+1)-(2y+x)}.`

I find this soln. of mine very easier than the one I provided earlier!