# Tangent To The Curve

1. Sep 9, 2006

### 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.

2. Sep 9, 2006

### 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?

3. Sep 9, 2006

### HallsofIvy

Staff Emeritus
The two equations you cite are in fact the same equation. The only difficulty is finding y'. As vsage says, use implicit differentiation.
If 2exy - 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'.

4. Sep 9, 2006

### E=m(C)^2

Oh ofcourse, seems so simple now. Thanks alot guys.