Tangent To The Curve

  • Thread starter E=m(C)^2
  • Start date
  • #1
14
0
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
IMG00003.GIF
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.
 

Answers and Replies

  • #2
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
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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
14
0
Oh ofcourse, seems so simple now. Thanks alot guys.
 

Related Threads on Tangent To The Curve

Replies
1
Views
7K
Replies
7
Views
1K
Replies
4
Views
9K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
Top