Solving Tangent Line Problem: Adding & Subtracting 5 Explained

[Bryanaam]

Homework Statement

Ok, how do i solve this question
I've attached the picture.. The answer is A but i don't understand how to get that...

The Attempt at a Solution

The solution is also given , but i don't understand the adding and subtracting 5. the Derivative of

y=ln (e^x + 5) y'= 1/ ( e ^ x + 5) , where is the extra e^x coming from in the first line of the solution. Also i already mention the adding and substracing 5.. I have no idea what's going on .. PLz Math guru's please help me!

 

[CompuChip]

y=ln (e^x + 5) y'= 1/ ( e ^ x + 5) , where is the extra e^x coming from in the first line of the solution.

You have to use the chain rule: y = ln(u) with u = e^x + 5; then y' = dy/du * du/dx

For the rest of your question we'll have to wait to see the attachments.

 

[HallsofIvy]

As CompuChip said, the "extra" e^x is from the chain rule:
ln(e^x+ 5)= ln(u) with u= e^x+ 5. dy/dx= (dy/du)(du/dx)
dy/du= d(ln(u))/du= 1/u and du/dx= d(e^x+ 5)/dx= e^x
dy/dx= (dy/du)(du/dx)= (1/u)(e^x)= (1/(e^x+5))e^x)= e^x/(e^x+ 5).

The first attachment asks what happens to the slope of the tangent lines to y= ln(e<sup>x[/sup+ 5) as x increases and answers that it increases to a limit of 1.

The second explains that answer by noting that y'= ex/(ex+ 5), as above, and then does the following:
1) add and subtract 5
[tex]\frac{e^x}{e^x+ 5}+ 5- 5[/tex]
which is incorrectly shown!

Probably whoever did the "art" did not understand what was meant. It should be
[tex]\frac{e^x+ 5- 5}{e^x+ 5}[/tex]
with the "5, added and subtracted" in the numerator of the fraction. Of course, since 5- 5= 0, those are equal. Now separate into two fractions
[tex]\frac{e^x+ 5}{e^x+ 5}- \frac{5}{e^x+ 5}= 1- \frac{5}{e^x- 5}[/tex]
As x increases without bound, so does ex and so does the denominator of the second fraction: the fraction goes to 0 so the entire expression goes to 1.

I would have done it a slightly different way, myself. From the original expression,
[tex]\frac{e^x}{e^x+ 5}[/tex]
instead of adding and subtracting 5, divide both numerator and denominator by ex:
[tex]\frac{\frac{e^x}{e^x}}{\frac{e^x+ 5}{e^x}}= \frac{1}{1+ 5e^{-x}}[/tex]
Now, as x increases without bound, 5e-x goes to 0 so that fraction goes to 1.</sup>

 

[CompuChip]

And a final way would probably to use L'Hopitals rule, but I will not explain that unless the OP requests it because HallsofIvy shows the most common ways, apart from some general techniques with fractions which are quite important.

 

[Bryanaam]

Thankyou so much Guys.. I think i got the idea now.. BTW Compuchip , what's this L'Hopital rule and how is it applied to this question. I'm curious to find out.. Is it a faster method for doing this problem?

Thanks in Advance.. You guys are awesome

 

[CompuChip]

L'Hopitals rule can be used if you have to evaluate a limit
[tex]\lim_{x \to a} \frac{f(x)}{g(x)}[/tex]
where f and g separately tend to either 0 or (plus or minus) infinity. In other words, if just plugging in x = a would produce something like 0/0, or [itex]\infty / \infty[/itex] (or [itex]\infty / -\infty[/itex], [itex]-\infty / \infty[/itex] or [itex]-\infty / -\infty[/itex]).

In this case, you may calculate
[tex]\lim_{x \to a} \frac{f'(x)}{g'(x)}[/tex]
and this is equal to original limit.

For your current problem, e^x blows up as x goes to infinity, so the limit would become of the form [itex]\infty / \infty[/itex].
Therefore, by L'Hopitals rule,
[tex]\lim_{x \to \infty} \frac{e^x}{e^x + 5} = \lim_{x \to \infty} \frac{(e^x)'}{(e^x + 5)'} = \lim_{x \to \infty} \frac{e^x}{e^x} = \lim_{x \to \infty} 1 = 1.[/tex]

However, I suggest one of the methods HallsofIvy has shown in this case.