umm i don't remember how to find the slope for this problem.

use implicit differentiation to find and equation of the line tangent to the graph at the point (2,0)

e^y + x^2 - 3y^2=5

so i know the tangent formula is (y-y1) = m (x-x1)

Printable View

- Mar 25th 2010, 12:36 PMFinalFantasy9291implicit differentiation
umm i don't remember how to find the slope for this problem.

use implicit differentiation to find and equation of the line tangent to the graph at the point (2,0)

e^y + x^2 - 3y^2=5

so i know the tangent formula is (y-y1) = m (x-x1) - Mar 25th 2010, 12:40 PMGeneral
**Well,**

**you have x1 = 2 , y1 = 0 ..**

**By differentiating both sides w.r.t. x, we got:**

**$\displaystyle y' e^y + 2x - 6yy' = 0 $**

**solve for y' , then evaluate it at (2,0), i.e. when x=2 and y=0 to get the value of m ..**

**then substitute x1,y1 and m the tangent line's formula ..**

**Please, note that its called "tangent line's formula" not "tangent formula" ..** - Mar 25th 2010, 12:47 PMakilele
..

- Mar 25th 2010, 12:49 PMFinalFantasy9291
thank you for that correction haha.

so from what you gave me i got (y-0)=-4(x-2) is that correct? - Mar 25th 2010, 12:51 PMGeneral
- Mar 25th 2010, 12:59 PMFinalFantasy9291
okay so..

y'e^y+2x-6yy' = 0

btw is there a way i can type this out without having to use ^ so it's not as confusing for you or me?

y'e^0+(2)-6(0)y'=0

y' + 4= 0

y'=-4

so... (y-0)=-4(x-2) - Mar 25th 2010, 01:01 PMGeneral
- Mar 25th 2010, 01:31 PMFinalFantasy9291
hehe. how do i use the latex?

tyvm - Mar 25th 2010, 01:40 PMGeneral