# I stumbled upon this calculus implicit differential question: find (du)/(dy) of the function u=sin(y^2+u). The answer is (2ycos(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.

I stumbled upon this calculus implicit differential question: find $\frac{du}{dy}$ of the function $u=\mathrm{sin}\left({y}^{2}+u\right)$.
The answer is $\frac{2y\mathrm{cos}\left({y}^{2}+u\right)}{1-\mathrm{cos}\left({y}^{2}+u\right)}$. 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.
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ordonansexa
We have $u=\mathrm{sin}\left({y}^{2}+u\right)$, differentiating gives
$\begin{array}{r}\frac{du}{dy}=\left(2y+\frac{du}{dy}\right)\mathrm{cos}\left({y}^{2}+u\right)\end{array}$
This can be rearranged to
$\begin{array}{r}\frac{du}{dy}\left(1-\mathrm{cos}\left({y}^{2}+u\right)\right)=2y\mathrm{cos}\left({y}^{2}+u\right).\end{array}$
So
$\begin{array}{r}\frac{du}{dy}=\frac{2y\mathrm{cos}\left({y}^{2}+u\right)}{1-\mathrm{cos}\left({y}^{2}+u\right)}.\end{array}$
###### Did you like this example?
Kathy Guerra
When applying the chain rule, you need the differentiate $u$ also. Explicitly,
$\frac{\mathrm{d}u}{\mathrm{d}y}=\frac{\mathrm{d}}{\mathrm{d}y}\left(\mathrm{sin}\left({y}^{2}+u\right)\right)=\mathrm{cos}\left({y}^{2}+u\right)\left(\frac{\mathrm{d}}{\mathrm{d}y}\left({y}^{2}+u\right)\right)=\mathrm{cos}\left({y}^{2}+u\right)\left(2y+\frac{\mathrm{d}u}{\mathrm{d}y}\right).$
Solving for $\frac{\mathrm{d}u}{\mathrm{d}y}$, we find the answer you announced.