# Thread: More derivatives (logarithmic)... due tomorrow

1. ## More derivatives (logarithmic)... due tomorrow

I got most of the worksheet done by myself (thanks to some browsing on these forums), but the last 3 are giving me some trouble. I got:

1.

I feel like I could possibly do this one, but I'm not sure how the algebraic rule of 1/x affects this problem. Do we have to do some type of conversion before we actually start taking the derivative and breaking down the problem?

2.

Raising the fraction to the power of x is really tripping me up on this one. Not sure where to go with it to be honest.

3.

Another fraction one, just like the 1st one.

I'm been really grinding at this thing now and been trying to get my derivatives down, but these last 3 are really breaking my back. The worksheet is due tomorrow morning, so some quick help would be much appreciated.

2. what he said down there, wow.. now I remember that, jeez talk about new material replacing old material.. thanks for taking me through a lesson too, soroban *thumbs up*

3. Hello, Drakeman!

We need to take logs first . . .

$1)\;\;\lim_{x\to0}\,\left(e^x + x\right)^{\frac{1}{x}}$
Let $y \;=\;(e^x + x)^{\frac{1}{x}}$

Take logs: . $\ln(y) \;=\;\ln(e^x+x)^{\frac{1}{x}}\;=\;\frac{1}{x}\!\cd ot\!\ln(e^x+x) \;=\;\frac{\ln(e^x+x)}{x}$

Take the limit: . $\lim_{x\to0}\,\bigg[\ln(y)\bigg] \;=\;\lim_{x\to0}\left[\frac{\ln(e^x+x)}{x}\right] \to \;\frac{0}{0}$

Apply L'Hopital: . $\lim_{x\to0}\,\bigg[\ln(y)\bigg] \;=\;\lim_{x\to0}\left[\frac{\frac{e^x+1}{e^x+x}}{1}\right] \;=\;\lim_{x\to0}\frac{e^x+1}{e^x+x}\;=\;\frac{1+1 }{1+0} \;=\;2$

Since $\lim_{x\to0}\,\bigg[\ln(y)\bigg] \:=\: 2$, then: . $\lim_{x\to0} y \:=\: e^2$

4. 2.

$lim_{x\to0}\ , \ (1/x^2)^x$
$lim_{x\to0}\ , \ ln(1/x^2)^x$
$lim_{x\to0}\ , \ ln(1/x^2)^x$
$lim_{x\to0}\ , \ (ln 1 - ln(x^2)^x$

as $lim_{x\to0}\ , \ = infinity$

therefore the limit does not exist

since you cant take ln0