Find $\frac {dy}{dx}$
For x = $\frac {2y}{x^2 - y}$
I have solved this in the easy route of multiplying by (x^2 - y).
The final answer is: $\frac {dy}{dx}$ = (3x^2 - y) / (x+2)
I want to solve using the quotient rule, directly but I can't do it. Here is my solution/working out.
https://i.stack.imgur.com/ay8aX.jpg
Thnx x
On
Differentiating with respect to $x$ we have by the quotient rule: $$1=\frac{2\frac{\mathrm d y}{\mathrm d x}(x^2-y)-(2x-\frac{\mathrm d y}{\mathrm d x})2y}{(x^2-y)^2}$$ so rearranging we have $$\frac{\mathrm d y}{\mathrm d x}=\frac{(x^2-y)^2+4xy}{2x^2}=\frac{(x^2-y)^2+4xy}{4\frac y x + 2y}=\frac{x(x^2-y)^2+4x^2y}{2y(x+2)}=\frac{4\frac{y^2}x+4x^2y}{2y(x+2)}=\frac{2\frac y x + 2x^2}{x+2}=\frac{3x^2-y}{x+2}$$ as required.
$$ x' = \frac{2y'}{x^2-y} - \frac{2y(2x x' - y')}{(x^2-y)^2} = \frac{2y'-x(2xx'-y')}{x^2-y} = \frac{(x+2)y'-2x^2x'}{x^2-y}, $$ replacing the extra $2y/(x^2-y)$. So $$ \frac{dy}{dx} = \frac{x^2-y+2x^2}{x+2} $$ and so on.
It's certainly not that obvious where to replace the values to get the given form: there are plenty of alternatives, not least solving for $y$ directly to get $$ y = \frac{x^3}{x+2} $$ and just differentiating that.