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

2 Answers

#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!