# Tangent Line Question

• Oct 31st 2006, 04:58 PM
drewms64
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?
• Oct 31st 2006, 06:35 PM
topsquark
Quote:

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?

$\frac{(3x+3h+5)e^xe^h - (3x+5)e^x}{h}$

I presume you are calculating the first derivative longhand?

I would actually rewrite the second line as:

$\frac{(3x+5)e^xe^h - (3x+5)e^x + 3he^xe^h}{h}$

$(3x+5)e^x \left ( \frac{e^h-1}{h} \right ) + 3e^xe^h$

When you take your limit as h goes to 0 the second term is easy. What you have left is to figure out what
$\lim_{h \to 0} \frac{e^h-1}{h}$

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

-Dan
• Oct 31st 2006, 07:01 PM
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
• Oct 31st 2006, 07:34 PM
topsquark
Quote:

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.

Quote:

Originally Posted by topsquark
$\frac{(3x + 3h + 5)e^xe^h - (3x+5)e^x}{h}$
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.

$\frac{(3x+5)e^xe^h - (3x+5)e^x + 3he^xe^h}{h}$

$\frac{ (3x+5)e^x (e^h - 1) + 3he^xe^h}{h}$ <-- Here's the factorization.

Now I'm going to split the fraction into two terms and divide by h:
$(3x+5)e^x \left ( \frac{e^h-1}{h} \right ) + 3e^xe^h$

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
• Oct 31st 2006, 08:34 PM
drewms64
Quote:

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.