Thread: Tangent Line Question

Hi, I have a quick question about finding the equation of a tangent line to the graph of y = (3x+5)e^x at x=0

so far I have Point (0,5)

Code:

(3(x+h) +5)e^(x+h) - (3x+5)e^x
--------------------------------
               h

(3x+3h+5)e^x + e^h - (3x+5)e^x
--------------------------------
               h

would I factor for the next step?

Originally Posted by drewms64

Hi, I have a quick question about finding the equation of a tangent line to the graph of y = (3x+5)e^x at x=0

so far I have Point (0,5)

Code:

(3(x+h) +5)e^(x+h) - (3x+5)e^x
--------------------------------
               h

(3x+3h+5)e^x + e^h - (3x+5)e^x
--------------------------------
               h

would I factor for the next step?

To start with you have an error in your second line. It should read:



I presume you are calculating the first derivative longhand?

I would actually rewrite the second line as:





When you take your limit as h goes to 0 the second term is easy. What you have left is to figure out what


For the record, the limit is 1. If you need help with it, just let me know.

-Dan

Ah thanks I missed that error, also are there any other ways to go from my second step onward? I'm just unsure of the way you did it

Edit: such as maybe

Code:

3xe^h + 3he^h +5e^h
----------------------
              h

although im not sure if that would work

Originally Posted by drewms64

Ah thanks I missed that error, also are there any other ways to go from my second step onward? I'm just unsure of the way you did it

Edit: such as maybe

Code:

3xe^h + 3he^h +5e^h
----------------------
              h

although im not sure if that would work

Factoring the e^x out of the whole expression would also work. It's just a matter of taste.

Originally Posted by topsquark

To start with you have an error in your second line. It should read:

What I am doing between these lines is simply rearranging the terms so I can factor a (3x+5)e^x from the first two terms.



<-- Here's the factorization.

Now I'm going to split the fraction into two terms and divide by h:

Hope that helps. Again the explicit way you break the problem up doesn't matter. You'll get the same result in the end.

-Dan

Originally Posted by topsquark

Factoring the e^x out of the whole expression would also work. It's just a matter of taste.

Yeah that helps showing me how you did it your way. If I was to do it by factoring out the e^x, how would I go about canceling out the H's from the e^h's? This way I can plug in 2 to find the slope to get the formula for the tangent line.