# finding the slope of the tangent line

• Nov 29th 2009, 03:45 PM
johnnyc418
finding 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?
• Nov 29th 2009, 05:36 PM
mamen
Quote:

Originally Posted by johnnyc418
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?

if you'll find/get the derivative of f(x)=e^(-x )+ e^(2x-1) and look for f'(1);that is,for example, f'(1)=n, then n is the slope of the tangent line at x=1.
• Nov 29th 2009, 06:32 PM
johnnyc418
just to verify, the derivative is y'=-e^-x + 2 correct? so then i would do
y'=-e^-1 +2?
• Nov 29th 2009, 06:38 PM
Jameson
Quote:

Originally Posted by johnnyc418
just to verify, the derivative is y'=-e^-x + 2 correct? so then i would do
y'=-e^-1 +2?

If y=e^(-x)+e^(2x-1) then no that's not the derivative. The second term is wrong. It should be (2)e^(2x-1). Do you see why?

Once you get the right y', plug in x=1 to get the slope of y at x=1. Then use the point-slope formula: $\displaystyle y-y_0=m(x-x_0)$ to find the tangent line.
• Nov 29th 2009, 07:01 PM
johnnyc418
Quote:

Originally Posted by Jameson
If y=e^(-x)+e^(2x-1) then no that's not the derivative. The second term is wrong. It should be (2)e^(2x-1). Do you see why?

Once you get the right y', plug in x=1 to get the slope of y at x=1. Then use the point-slope formula: $\displaystyle y-y_0=m(x-x_0)$ to find the tangent line.

no not really. i dont understand how youre getting (2)e^(2x-1)
• Nov 30th 2009, 12:16 AM
Jameson
Quote:

Originally Posted by johnnyc418
no not really. i dont understand how youre getting (2)e^(2x-1)

$\displaystyle \frac{d}{dx}e^{f(x)}=e^{f(x)}*f'(x)$

It's the chain rule property used so often. The derivative of (2x-1) is just 2, so that is why there is a 2 in front now. The exponent of e remains the same.
• Nov 30th 2009, 01:52 AM
johnnyc418
Quote:

Originally Posted by Jameson
$\displaystyle \frac{d}{dx}e^{f(x)}=e^{f(x)}*f'(x)$

It's the chain rule property used so often. The derivative of (2x-1) is just 2, so that is why there is a 2 in front now. The exponent of e remains the same.

what about the first part that says e^-x. isnt that equal to -e^-x? what happens to that?
• Nov 30th 2009, 01:55 AM
Jameson
Quote:

Originally Posted by johnnyc418
what about the first part that says e^-x. isnt that equal to -e^-x? what happens to that?

I was only referencing the error you made, which was in the 2nd term. Sorry if that was confusing. You are correct that the derivative of e^(-x) is -e^(-x)
• Nov 30th 2009, 02:14 AM
johnnyc418
so would the derivative be y'=-e^-x + (2)e^(2x-1)?
• Nov 30th 2009, 02:17 AM
Jameson
Quote:

Originally Posted by johnnyc418
so would the derivative be y'=-e^-x + (2)e^(2x-1)?

Yes.