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I've been stuck on a certain implicit differentiation problem that I've tried several times now. $$ \frac{x^2}{x+y} = y^2+6 $$

I know to take the derivatives of both sides and got: $$ \frac{(x+y)2x-\left(1-\frac{dy}{dx}\right)x^2}{(x+y)^2} = 2y\frac{dy}{dx} $$

I reduced that to get:

$$2x^2 +2xy-x^2-x^2*dy/dx=(2y*dy/dx)(x+y)^2$$

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 factored 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:

$$dy/dx=(2y(x+y)^2)/(4x^7y)$$

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!

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  • $\begingroup$ just after the line : i reduced that to get , you have +x^2 instead of -x^2. $\endgroup$ – hamam_Abdallah Sep 28 '16 at 21:17
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An idea to avoid the cumbersome and annoying quotient rule: multiply by the common denominator

$$\frac{x^2}{x+y}=y^2+6\implies xy^2+6x+y^3+6y-x^2=0\implies$$

$$y^2+2xyy'+6+3y^2y'+6y'-2x=0\implies(2xy+3y^2+6)y'=2x-y^2-6\implies$$

$$y'=\frac{2x-y^2-6}{2xy+3y^2+6}$$

If nevertheless you want to use the quotient rule:

$$\frac{2x(x+y)-x^2}{(x+y)^2}-\frac{x^2}{(x+y)^2}y'=2yy'\implies$$

$$(2y(x+y)^2+x^2)y'=x^2+2xy\implies y'=\frac{x^2+2xy}{2y(x+y)^2+x^2}$$

Now, how come both expressions we got are equal?. Well, for one we can use, for example, that $\;\cfrac{x^2}{x+y} = y^2+6\;$ , so

$$\frac{x^2+2xy}{2y(x+y)^2+x^2}=\frac{2x-y^2-6}{2xy+3y^2+6}=\frac{2x-\frac{x^2}{x+y}}{2xy+2y^2+\frac{x^2}{x+y}}\iff$$

$$\frac{x^2+2xy}{2y(x+y)^2+x^2}=\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}\iff$$

$$\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}=\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}\;\;\color{green}\checkmark$$

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