<s>[INST] Find derivative of $u = \sin(y^2 + u)$ using implicit differentiation Find $\frac{du}{dy}$, when $u = \sin(y^2 + u)$, using chain rule and expanding I get:
$1 = (2y\frac{du}{dy} + 1) \cos(y^2 + u)$
$2y \cos(y^2 + u)\frac{du}{dy} + \cos(y^2 + u) = 1$ 
$2y \cos(y^2 + u)\frac{du}{dy} = 1- \cos(y^2 + u)$
$\frac{du}{dy} = \frac{1 - \cos(y^2 + u)}{2y \cos(y^2 + u)}$
Since this was wrong, I massaged it a bit more to:
$\frac{du}{dy} = \frac{\sec(y^2 + u) - 1}{2y}$
Still wrong... But what am I doing wrong, wolfram alpha seems to agree with me...
 [/INST] Taking derivative towards $u$ w.r.t.$y$ is not $1$, but should be $\frac{d u}{d y}$
Thus, it instead should be:
$$\frac{du}{dy}=\cos(y^2 + u)(2y + \frac{du}{dy})$$
$$\frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2 + u)}$$
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