Equation of Tangent to Curve 2e^(xy)-ysinx=log(y)+2 at (0,1)

[E=m(C)^2]

Hi guys, I'm having a little trouble finding the equation of the tangent to the curve 2e^(xy) - ysinx = log(y) + 2 at the point (0,1).
I've basically concluded that either the equation

should be used or either find the gradient at (0,1) through differentiation and then use (y-y1)=m(x-x1).
But i can't seem to get the derivative in terms of y. Any help or advice would be much appreciated, thank you.

 

[vsage]

I think the point was that you can't get the equation in terms of a solvable variable. You can, however, probably implicitly differentiate the above equation with respect to y and with respect to x. Have you tried that path yet?

 

[HallsofIvy]

The two equations you cite are in fact the same equation. The only difficulty is finding y'. As vsage says, use implicit differentiation.
If 2e^{xy - ysinx = log(y) + 2 then
2exy(y+ xy')- y' cos x- y sin x= y'/y. Set x= 0, y= 1 and solve for y'.}

 

[E=m(C)^2]

Oh ofcourse, seems so simple now. Thanks a lot guys.