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Find the derivative $$\frac{dy}{dx}$$ of the function $$y=\sin(x^2+y)$$ what rule should I use here? chain rule?
so, $$m=x^2+y$$ and $$(m)'= 2x+y \frac{dy}{dx}$$
$$(\sin(x^2+y))'= \cos(m) * m'$$
$$=\cos(x^2+y) (2x+y \frac{dy}{dx})$$
and what to do with y from the left side from the initial equation $$y=\sin(x^2+y)$$, does it's derivative will be 1?
so, $$1=\cos(x^2+y) (2x+y \frac{dy}{dx})$$
could anyone tell me please, if I am on a right way or wrong

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  • $\begingroup$ The derivative of $y$ with respect to $x$ is $\frac {dy}{dx}$ $\endgroup$ – Mark Bennet Jun 20 '15 at 16:38
  • $\begingroup$ the derivative of $m$ is just $2x+\frac{dy}{dx}$ $\endgroup$ – user210387 Jun 20 '15 at 16:40
  • $\begingroup$ $(m)'=2x+\frac{dy}{dx}$ $\endgroup$ – user31415926535 Jun 20 '15 at 16:41
  • $\begingroup$ I edited your question a little, replacing your "sin" etc. with the more $LaTeX$-friendly "\sin" and so forth; $\LaTeX$ has a number of built-ins for trig and other important functions. Cgeers! $\endgroup$ – Robert Lewis Jun 20 '15 at 16:46
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$y=\sin(x^2+y)$

$\dfrac{dy}{dx}=\cos(x^2+y)\Bigl(2x+\dfrac{dy}{dx}\Bigr)$

$\dfrac{dy}{dx}=\dfrac{2x\cos(x^2+y)}{1-\cos(x^2+y)}$

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  • $\begingroup$ I think it should be 2x $\endgroup$ – user210387 Jun 20 '15 at 16:42
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In the third line, you should have

$$m'=2x+\frac{dy}{dx}$$

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