Find the derivative of $y^3-xy^2+\cos xy=2$ Find the derivative of $y^3-xy^2+\cos xy=2$
My Attempt:
$$y^3-xy^2+\cos xy=2$$
$$\dfrac {d}{dx} [y^3-xy^2+\cos xy]=\dfrac {d}{dx} [2]$$
$$3y^2.\dfrac {dy}{dx} -[1.y^2+2xy\dfrac {dy}{dx}] + (-\sin xy) \dfrac {dy}{dx}(xy)=0$$
$$3y^2 \dfrac {dy}{dx} - y^2 - 2xy\dfrac {dy}{dx} - \sin xy (y+\dfrac {dy}{dx} x)=0$$
How do I proceed further?
 A: \begin{align}
y^3 - xy^2 + \cos xy=& 2\\
3y^2 \frac{dy}{dx} - y^2 - 2yx\frac{dy}{dx}-\sin xy(y+x\frac{dy}{dx}) =& 0\\
3y^2 \frac{dy}{dx} - y^2 - 2yx\frac{dy}{dx}-y\sin xy-x\sin xy\frac{dy}{dx} =& 0\\
\frac{dy}{dx}(3y^2 - 2xy - x\sin xy) =&y^2+y\sin xy
\end{align}
Hence
$$\frac{dy}{dx} = \frac{y^2+y\sin xy}{3y^2 - 2xy - x\sin xy}$$
A: One note, you will frequently get cleaner looking results if you take all of the derivatives with respect for x and all of the derivatives with respect to y separately.
That is:
$\frac {d}{dx} (y^3- xy^2 + \cos xy) = ( -y^2 - y \sin xy) + (3y^2-2xy - x\sin y)\frac {dy}{dx}$
Play with it and see that you get the same results as when you work left to right.
Either way, your goal at this point is to isolate the $\frac {dy}{dx}$.
$( -y^2 - y \sin xy) + (3y^2-2xy - x\sin xy)\frac {dy}{dx} = 0\\
\frac {dy}{dx} = \frac {y^2 + y\sin xy}{3y^2 - 2xy-x\sin xy}$
A: Collect all terms with $\dfrac{dy}{dx}$ on one side and everything else on the other side so you have something in this form:
$$P \cdot \frac{dy}{dx} = Q,$$
where $P$ and $Q$ will be some expressions involving $x$ and $y$.  Then divide both sides by $P$.
Note that to get started on these next steps you'll need to first distribute that $-\sin xy$ factor into the $\left(y + \dfrac{dy}{dx} x\right)$.
A: So your last line is:
\begin{align}
3y^2\dfrac{dy}{dx}-y^2-2xy\dfrac{dy}{dx}-\left(y+\dfrac {dy}{dx}x\right)\sin (xy)&=0\\
\implies\left(3y^2-2xy-x\sin(xy)\right)\dfrac{dy}{dx}&=y^2+y\sin(xy)\\
\implies\dfrac{dy}{dx}&=\dfrac{y^2+y\sin(xy)}{3y^2-2xy-x\sin(xy)}
\end{align}
A: Generally,  the derivative of close function is $$F\left( x,y \right) =0\\ d\left( F\left( x,y \right)  \right) =0\\ { F }_{ x }^{ \prime  }\left( x,y \right) dx+{ F }_{ y }^{ \prime  }\left( x,y \right) dy=0\\ \frac { dy }{ dx } =-\frac { { F }_{ x }^{ \prime  }\left( x,y \right)  }{ { F }_{ y }^{ \prime  }\left( x,y \right)  } $$
