Test review for calculus, related rates & critical numbers I got 12/60 on my latest test I want to go over it to figure out what I did wrong and why and how to improve. I was looking if there is anything specific, any pattern to this. When I review my tests I just see all sorts of problems, I forget formulae, I forget basic math, I mess up signs, I run out of time and can't finish the test. Things like that, I don't have an answer key for this test is there any way I can go over it on here and someone else can tell me why I am messing up so much?
There 10 questions and I could probably type them up in under 30 minutes, and another 30 for my work.
Anyways here is the first one. I am supposed to find the derivative of $\log 3 (x+ \sqrt {(x^2-1})$.
I then set the problem up as $1/u \ln 3 (1+1/2(x^2-1)^{-1/2}2x)$; this is wrong, completely wrong.
Next questions is let $f(x) = (\cos x)^x$ and find the derivative.
I couldn't even do this on the test, I forgot how to get the derivative of a variable that is an exponent, I knew what to do I just couldn't make sense of how to do and I have scrawled down about 30 attempts at nothing.
Doing this now I think what I need to do is rewrite the problem as $e^{\ln u^x}$ and then solve it from there. I could be wrong though.
3) I am supposed to solve this word problem. 

A particle moves along a real axis and its location, $s$, is given by $s=t^3-12t^2+36t$ ($s$=feet)
  a) find the acceleration at $t=2$ seconds

For this I know that I need to get the derivative and put in $2$ for $t$. Pretty simple, I makes the problem $3t^2-24t+36=0$; I think I forgot to do this one, not sure what happened but this is all I have written down, not sure why. I would plug in $2$ and get $0$, oh I guess I did have the answer written down. It was probably just wrong.

b) When is the particle moving in the positive direction?

I know for this I need to find the critical numbers and test points around them to see where they are increasing or decreasing. I have positive from $5$ to $\infty$. 

c) What is the total distance traveled in the first $6$ seconds? 

I have written down $39$ ft.
I am not sure what the rules on this are so I will not post anymore questions under this topic unless I know it is okay to. I would like to transcribe the whole test if that is possible.
 A: When you have something of the form $f(x) = x^x$, you need to do something like this:
$$f(x) = x^x$$
$$\ln f(x) = \ln(x^x)$$
$$\ln f(x) = x \ln x$$
Then differentiate both sides:
$$\frac{f'(x)}{f(x)} = \ln x + 1$$
$$f'(x) = x^x(\ln x + 1)$$
It should be simple to figure out what to do when $f(x) = (\cos x)^x$. Just use the chain rule in combination with the above.
Regarding problem 3a), as it was said in another answer, the acceleration is the second derivative of the position. In this case, $a = 6t - 24$.

I agree with the commenters. I think you should get a tutor. If tutors are too expensive, maybe ask a friend for help. It sounds like you need to go over some of the basics, and just getting the answer to your problems here isn't going to be of much help.
A: For the first question:
$$\begin{align*}
f(x) &= \log_3(x + \sqrt{x^2 - 1})\\  
f'(x) &= \frac{1}{\ln3}\frac{1}{x + \sqrt{x^2 - 1}}\left(1 + \frac{1}{2}\frac{1}{\sqrt{x^2 - 1}}2x\right)\\  
f'(x) &= \frac{1}{\ln3}\left(\frac{1}{x + \sqrt{x^2 - 1}} + \frac{x}{(x + \sqrt{x^2 -1})\sqrt{x^2 -1}}\right)\\  
f'(x) &=\frac{1}{\ln3}\left(\frac{x + \sqrt{x^2 -1}}{(x + \sqrt{x^2 -1})\sqrt{x^2 -1}}\right)\\
f'(x) &=\frac{1}{\ln3}\frac{1}{\sqrt{x^2 -1}}
\end{align*}$$
A: For the 3a, the acceleration is the second derivative of the position, so you need to take one more derivative before plugging in $t=2$.  It is not equal to $0$.  I get $-12$ as the acceleration at $t=2$.  For 3b, your derivative is the velocity, so you can find its roots.  I find it positive over $(-\infty,2)\cup(6,\infty)$.  For 3c, the particle moves right until $t=2$, reaching $s=32$, then returns at $t=6$ to $s=0$, so it travels a total of $64$.
