I've been stuck on a certain implicit differentiation problem that I've tried several times now. (x^2)/(x+y)=y^2+6

kislotd

kislotd

Answered question

2022-07-22

I've been stuck on a certain implicit differentiation problem that I've tried several times now.
x 2 x + y = y 2 + 6
I know to take the derivatives of both sides and got:
( x + y ) 2 x ( 1 d y d x ) x 2 ( x + y ) 2 = 2 y d y d x
I blackuced that to get:
2 x 2 + 2 x y x 2 x 2 d y / d x = ( 2 y d y / d x ) ( x + y ) 2
I then divided both sides by (2y*dy/dx) and multiplied each side by the reciprocals of the first three terms of the left side. Then I factoblack dy/dx out of the left side and multiplied by the reciprocal of what was left to get dy/dx by itself. I ended up with:
d y / d x = ( 2 y ( x + y ) 2 ) / ( 4 x 7 y )
but this answer was wrong. I only have one more attempt on my online homework and I can't figure out where I went wrong. Please help!

Answer & Explanation

Jazlene Dickson

Jazlene Dickson

Beginner2022-07-23Added 15 answers

An idea to avoid the cumbersome and annoying quotient rule: multiply by the common denominator
x 2 x + y = y 2 + 6 x y 2 + 6 x + y 3 + 6 y x 2 = 0
y 2 + 2 x y y + 6 + 3 y 2 y + 6 y 2 x = 0 ( 2 x y + 3 y 2 + 6 ) y = 2 x y 2 6
y = 2 x y 2 6 2 x y + 3 y 2 + 6
If nevertheless you want to use the quotient rule:
2 x ( x + y ) x 2 ( x + y ) 2 x 2 ( x + y ) 2 y = 2 y y
( 2 y ( x + y ) 2 + x 2 ) y = x 2 + 2 x y y = x 2 + 2 x y 2 y ( x + y ) 2 + x 2
Now, how come both expressions we got are equal?. Well, for one we can use, for example, that x 2 x + y = y 2 + 6 , so
x 2 + 2 x y 2 y ( x + y ) 2 + x 2 = 2 x y 2 6 2 x y + 3 y 2 + 6 = 2 x x 2 x + y 2 x y + 2 y 2 + x 2 x + y
x 2 + 2 x y 2 y ( x + y ) 2 + x 2 = x 2 + 2 x y 2 x 2 y + 4 x y 2 + 2 y 3 + x 2
x 2 + 2 x y 2 x 2 y + 4 x y 2 + 2 y 3 + x 2 = x 2 + 2 x y 2 x 2 y + 4 x y 2 + 2 y 3 + x 2

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