# Find dy/dx by implicit differentiation $e^y\cos x=1+\sin(xy)$?

I would appreciate so much if someone would be willing to help me understand how to correctly solve this problem...

I need to find $\frac {dy} {dx}$: $$\frac{d} {dx} e^y\cos x=1+\sin(xy)$$

using implicit differentiation.

So far I've gotten to this point which I don't think I'm doing right because then I get stuck:

$$e^y(-\sin x)+(\cos x)(e^y)(\frac {dy} {dx})=\cos(xy)(x)(\frac {dy} {dx})+y(1)$$

Am I right? Am I wrong? If I'm right I definitely have no clue where to go...and if I'm wrong...then again I have no idea what to do.

• You are missing the brackets on the RHS. $e^y(-\sin x)+(\cos x)(e^y)(\frac {dy} {dx})=0+ \cos(x y)(x(\frac {dy}{dx})+ y.1).$ – Mhenni Benghorbal Feb 27 '13 at 19:52
• Mhenni is correct. From that point, collect $\frac{dy}{dx}$ terms on one side, factor it out, and divide as in Kaster's answer. – Mike Feb 27 '13 at 19:56

$$e^y \cos x = 1 + \sin xy \\ e^y y' \cos x - e^y \sin x = \cos xy (y + xy') \\ y' (e^y \cos x-x\cos xy) = y \cos xy+e^x \sin x \\ y' = \frac {y \cos xy+e^x \sin x}{e^y \cos x-x\cos xy}$$
$$e^y\cos x=1+\sin(xy)$$ $$e^yy'\cos x-\sin xe^y=\cos(xy)(y+xy')$$