How do you implicitly differentiate -1=(x+y)^2-xy-e^(3x+7y) โˆ’1=(x+y)2โˆ’xyโˆ’e3x+7y?

2 Answers
Jun 24, 2016

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}.

Jun 24, 2016

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!