i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?

Printable View

- November 29th 2009, 04:45 PMjohnnyc418finding the slope of the tangent line
i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?

- November 29th 2009, 06:36 PMmamen
- November 29th 2009, 07:32 PMjohnnyc418
just to verify, the derivative is y'=-e^-x + 2 correct? so then i would do

y'=-e^-1 +2? - November 29th 2009, 07:38 PMJameson
- November 29th 2009, 08:01 PMjohnnyc418
- November 30th 2009, 01:16 AMJameson
- November 30th 2009, 02:52 AMjohnnyc418
- November 30th 2009, 02:55 AMJameson
- November 30th 2009, 03:14 AMjohnnyc418
so would the derivative be y'=-e^-x + (2)e^(2x-1)?

- November 30th 2009, 03:17 AMJameson