1. ## implicit 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)

2. Originally Posted by FinalFantasy9291
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)
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" ..

3. ..

4. thank you for that correction haha.

so from what you gave me i got (y-0)=-4(x-2) is that correct?

5. Originally Posted by FinalFantasy9291
thank you for that correction haha.

so from what you gave me i got (y-0)=-4(x-2) is that correct?
Let me see your work ..
mayba you have mistakes ..
If you got the right final answer, that does not mean you've solved the problem correctly ..

6. 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)

7. Originally Posted by FinalFantasy9291
btw is there a way i can type this out without having to use ^ so it's not as confusing for you or me?
Yes. Use the latex.

Originally Posted by FinalFantasy9291
o
y'e^0+(2)-6(0)y'=0
I think that 2 is just a typo .. Right?
It should be 4..

Originally Posted by FinalFantasy9291
so... (y-0)=-4(x-2)
Correct.

8. hehe. how do i use the latex?

tyvm

9. Originally Posted by FinalFantasy9291
hehe. how do i use the latex?

tyvm
http://www.mathhelpforum.com/math-help/latex-help/