# Find the tangent line of $\frac{x^2}{y+1}+xy^2=4$ at $y=1$ and where $y<x$

I want to find the tangent line for the function $\frac{x^2}{y+1}+xy^2=4$ at the point $y=1$ and where $y<x$.

First step: finding the point so I inserted y=1 and get : $$x^2+2x-8 \rightarrow x_1=2 ,x_2=-4$$ $x_1 = 2$ satisfies the conditions.

Step two: finding the derivative $$\frac{2x(y+1)-y'x^2}{(y+1)^2}+2xy\times y'= 0$$ $$2x(y+1)-y'x^2+2xyy'(y+1)^2=0 \rightarrow y'(-x^2+2xy(y+1)^2)=-2x(y+1)$$ $$y'=\frac{-2x(y+1)}{-x^2+2xy(y+1)^2}$$ What I get after set $y=1,x=2$ is $-\frac{8}{12}$ and the answer is $-\frac{12}{12}$ so I guess I went wrong in the derivation process. I may have a fundamental error of the implicit function derivation? I`d like to get some advice.

Thanks.

You have forgotten that the $x y^2$ term requires a product rule when differentiating
• $xy^2$ is $y^2+2xy*y'$? – Ofir Attia Jun 2 '13 at 8:21