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...
 A: Taking derivative towards $u$ w.r.t.$y$ is not $1$, but should be $\frac{d u}{d y}$
Thus, it instead should be:
$$\frac{du}{dy}=\cos(y^2 + u)(2y + \frac{du}{dy})$$
$$\frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2 + u)}$$
A: i have $$u'=\cos(y^2+u)(2y+u')$$ therefore $$u'=\frac{2y\cos(y^2+u)}{1-\cos(y^2+u)}$$
A: You are almost there. Assuming that $u=u(y)$ and calling $u'=du/dy$ we have:
$$u'=(2y+u')\cos(y^2+u)\to u'=\frac{2y\cos(y^2+u)}{1-\cos(y^2+u)}$$
A: The expressions given so far can be said to suffer from the fact that the variables $y$ and $u$ are not separated. Although this was not requested in the OP we provide the solution with this additional feature.
The trick is to use $y$ as the dependent variable and $u$ as the independent one.
Solving
$$u = sin(u + y^2)$$
for $y$ gives
$$y = \pm \sqrt{\sin ^{-1}(u)-u}$$
Now
$$ \frac{\text{dy}}{\text{du}}=\pm\frac{\frac{1}{\sqrt{1-u^2}}-1}{2 \sqrt{\sin ^{-1}(u)-u}}$$
inverting gives finally
$$\frac{\text{du}}{\text{dy}}=\pm\frac{2 \sqrt{1-u^2} \sqrt{\sin ^{-1}(u)-u}}{1-\sqrt{1-u^2}}$$
This expression can be used, for example, to study the implicit function $u(y)$ close to $u = 0$. Which is found to be
$$u(y)\simeq \sqrt[3]{6 y^2}$$
