#x^2y + xy = 6#. Find #dx/dy#?
2 Answers
Explanation:
Treating
#=>d/dy (x^2y)+d/dy (xy)=d/dy(6)#
#=>[(d(x^2))/dy*y+x^2* (d(y))/dy]+[(d(x))/dy*y + x*(d(y))/dy]=0#
Let
#=>2x x'y+x^2(1)+x'y+x(1)=0#
#=>2x x'y+x'y+x^2+x=0#
#=>x'(2xy+y)="-"(x^2+x)#
#=>x'=-(x^2+x)/(2xy+y)=-(x(x+1))/(y(2x+1))#
Also:
#=>y(x^2+x)=6#
#=>y=6(x^2+x)^"-1"=6/(x(x+1))#
Substituting this into our equation for
#x'=-(x(x+1))/([6/(x(x+1))] (2x+1))#
#color(white)(x')=-(x^2(x+1)^2)/(6(2x+1))#
Another way:
We could also recognize that, since
#dy/dx=6("-"1)(x^2+x)^"-2"(2x+1)=-(6(2x+1))/((x^2+x)^2)#
And reciprocating this gives
#dx/dy=-(x^2+x)^2/(6(2x+1))=-([x(x+1)]^2)/(6(2x+1))=-(x^2(x+1)^2)/(6(2x+1))#
as before.
I got:
#(dx)/(dy) = -(x^2+x)/(2xy + y)#
A nice trick is to recognize that
#f(x,y) = x^2y + xy = 6#
From the chain rule,
#=> (x^2(d)/(dy) [y]cdot(dy)/(dx) + (d)/(dx) [x^2]y) + (x(d)/(dy) [y]cdot(dy)/(dx) + y(d)/(dx) [x]) = 0#
#=> (x^2(dy)/(dx) + 2xy) + (x(dy)/(dx) + y) = 0#
#=> x^2(dy)/(dx) + x(dy)/(dx) = -2xy - y#
#=> (dy)/(dx)[x^2 + x] = -(2xy + y)#
#=> (dy)/(dx) = -(2xy + y)/[x^2 + x]#
So,
Let's check the other way. If
#=> (x^2 (d)/(dy) [y] + y d/(dx) [x^2]cdot(dx)/(dy)) + (x(d)/(dy) [y] + y(d)/(dx) [x]cdot(dx)/(dy)) = 0#
#=> (x^2 + 2xy(dx)/(dy)) + (x + y(dx)/(dy)) = 0#
#=> 2xy(dx)/(dy) + y(dx)/(dy) = -x - x^2#
#=> (dx)/(dy)[2xy + y] = -(x + x^2)#
#=> color(blue)((dx)/(dy) = -(x + x^2)/[2xy + y])#
Indeed, same result both ways.