Find $\frac{dy}{dx}$ for $x^2=\frac{x-y}{x+y}$.
I have solved this in two ways.
First, I multiplicated the whole equation by $x+y$ and then I calculated the implicit derivative. I got the following solution:
$\frac{1-3x^2-2xy}{x^2+1}$
So far so good. When I calculated the implicit derivative of the original expression using the quotient rule though, I got a different solution, i.e.:
$-\frac{x(y+x)^2-y}{x}$
I have tried using Wolfram and I got the same results.
Can anyone explain to me why I get different solutions ?
Best Answer
The two solutions that you found are equal. So where is no contradiction.
Note that solving $\quad x^2=\frac{x-y}{x+y}\quad$ for $y$ gives $\quad y=x\frac{1-x^2}{1+x^2}$
Your first solution : $$\frac{1-3x^2-2xy}{x^2+1} =\frac{1-3x^2-2x\left(x\frac{1-x^2}{1+x^2} \right)}{x^2+1} = \frac{1-4x^2-x^4}{(1+x^2)^2}$$
Your second solution : $$-(x+y)^2+\frac{y}{x}=-\left(x+x\frac{1-x^2}{1+x^2}\right)^2+\frac{1}{x}\left(x\frac{1-x^2}{1+x^2}\right)=\frac{1-4x^2-x^4}{(1+x^2)^2}$$ Thus $$\frac{1-3x^2-2xy}{x^2+1}=-(x+y)^2+\frac{y}{x}$$