Calculus implicit differentiation question I stumbled upon this calculus implicit differential question: find
$\cfrac{du}{dy} $ of the function $ u = \sin(y^2+u)$.
The answer is $ \cfrac{2y\cos(y^2+u)}{1−\cos(y^2+u)} $.  I understand how to get the answer for the numerator, but how do we get the denominator part?
And can anyone point out the real intuitive difference between chain rule and implicit differentiation? I can't seem to get my head around them and when or where should I use them.
Thanks in advance!
 A: When applying the chain rule, you need the differentiate $u$ also. Explicitly,
$$
\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{\mathrm{d}}{\mathrm{d}y} \Big( \sin(y^2 + u) \Big) = \cos(y^2 + u) \Big( \frac{\mathrm{d}}{\mathrm{d}y} (y^2 + u) \Big) = \cos(y^2 + u) \Big( 2y + \frac{\mathrm{d}u}{\mathrm{d}y} \Big).
$$
Solving for $\frac{\mathrm{d}u}{\mathrm{d}y}$, we find the answer you announced.
A: We have  $ u=\sin(y^2+u)$, differentiating gives 
\begin{eqnarray*}
\frac{du}{dy} =\left( 2y+ \frac{du}{dy} \right) \cos(y^2+u) 
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
\frac{du}{dy}(1-\cos(y^2+u)) =2y \cos(y^2+u) .
\end{eqnarray*}
So
\begin{eqnarray*}
\frac{du}{dy} =\frac{ 2y \cos(y^2+u) }{1- \cos(y^2+u) }.
\end{eqnarray*}
A: $$u = \sin(y^2+u) \implies\frac{du}{dy} = \cos(y^2+u)\times\left(2y+\frac{du}{dy}\right)$$
$$\implies \frac{du}{dy} (1-\cos(y^2+u)) = 2y\cos(y^2+u) \implies \frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2+u)}$$
A: The answers above answer the first part of your question, but I'd also like to discuss the relationship between implicit differentiation and the chain rule.
Let's consider this example: x^2 + y^2 = 25
If we differentiate both sides, that is take d/dx on both sides of the equation, we are left with the following result: d/dx(x^2) + d/dx(y^2) = d/dx(25)
Using a few hopefully familiar rules, we may simplify the equation into:
2x + d/dx(y^2) = 0
The question we have is what exactly is d/dx(y^2)? To tackle this problem, we'll pretend that there is a relationship between y and x. That is, y = 'something' x
Assuming this relationship exists, we may now go onward. When we take d/dx(y^2) we are really taking d/dx(('something' x)^2). Using the chain rule, we find the answer to this must be 2 • ('something' x) • d/dx ('something' x). Substituting y for 'something' x we really just have 2y • dy/dx
Hope this helped!
A: Starting with:
$ u = \sin(y^2+u)$
Perhaps more apparent and facile path (especially on an exam where an equivalent valid solution may suffice) would be just take the inverse of the  sine function for both sides. This results here in a rapid and easy separation of $u$ and $y$:
$ \arcsin(u) = y^2+u $
$ \arcsin(u) - u = y^2 $
Taking the differentials of both sides and applying the formula for differentiating the arc sine function:
$ du/\sqrt(1-u^2) - du = 2 y dy $
Whereupon it is evident:
$ \cfrac{du}{dy} = \cfrac{2y}{1/\sqrt(1-u^2)-1} $
which I claim as a likely equivalent and facile solution path. Note: the domain of the inverse sin function can be likely defined in a suitable fashion to accommodate real world conditions (see comments here).
