Calculus implicit differentiation question

Solution 1:

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*}

Solution 2:

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.

Solution 3:

$$u = \sin(y^2+u) \implies\frac{du}{dy} = \cos(y^2+u)\times\left(2y+\frac{du}{dy}\right)$$ $$\implies \frac{du}{dy} (1-\cos(y^2+u)) = 2y\cos(y^2+u) \implies \frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2+u)}$$