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.
Best Answer
$$ e^y \cos x = 1 + \sin xy \\ e^y y' \cos x - e^y \sin x = \cos xy (y + xy') \\ y' (e^y \cos x-x\cos xy) = y \cos xy+e^x \sin x \\ y' = \frac {y \cos xy+e^x \sin x}{e^y \cos x-x\cos xy} $$