Question

http://www.ms.uky.edu/~ma123/
go to old exams, then exam 2 (latest one), and then problem 9

Answer 1

actually problem 10.

Answer 2

what rule is used there? product?

Answer 3

@saifoo.khan
@amistre64
@satellite73

Answer 4

You did the ranking wrong.
i should be the third one.

Answer 5

lol

Answer 6

WHere can i find the questions?

Answer 7

the link

Answer 8

Topic?

Answer 9

what?

Answer 10

its derivative stuff

Answer 11

Where should i go?
MA 123 Home
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Answer 12

umm i made it clear above

Answer 13

i still can't get where to go. lol

Answer 14

Old Exams
Exam 2 or Answer Key below it (which gives problems & answers)
Then problem 10

Answer 15

Now that's what i was talking about.

Answer 16

The answer's C.

Answer 17

Wanna know why?
let me send u a link

Answer 18

10 is D

Answer 19

http://ge.tt/1XHRScE/v/2?c

Answer 20

10 is C.

Answer 21

look at the answer key

Answer 22

OHHHHHHHHHHHHHHHHH!!!!
Sorry!
now i get the reason. $hit man!
they used Product rule

Answer 23

what about 11? what are the steps there
answers A btw

Answer 24

http://ge.tt/1XHRScE/v/2?c

Answer 25

It's mental problem man!

Answer 26

yeah i understand how they got what they got. what rule is used? i see that they took the derivative.

Answer 27

product rule

http://upload.wikimedia.org/wikipedia/en/math/9/c/e/9ceee641d053b28b97ca7f5023ec31ca.png

Answer 28

u sure? why is it a fraction then?

Answer 29

?

Answer 30

Let me show u the working. YOu are just confusing yourself bro.

Answer 31

lol i am

Answer 32

i also need 12 clarified as well

Answer 33

Question>?

Answer 34

11and12

Answer 35

Oh, Question 11 dosnt need any working.
it's simple.

Answer 36

Look, inside ln we had, 5x^2 +3x +7

Now what we have to do is,
write the inside thing as it is in the denominator.
And in the numerator, find the derivative of the inside thing.

Answer 37

something like,
\[\frac{\text{derivative of anything inside \ln}}{\text{anything inside \ln}}\]

Answer 38

getting something?

Answer 39

for the 12, the product rule was used right?

Answer 40

@myininaya
@JamesJ
@KingGeorge
You guys wanna join in? Only Saifoo has been helping me out.
I'm trying to understand this stuff so I can do well on my test.

Answer 41

For 12 it is the product rule.

Answer 42

what about 13?

Answer 43

13 * 10 =130

Answer 44

13 would be the chain rule. And I'm getting 130 as well.

Answer 45

yeah i have the answers. im just curious how they got each one.

Answer 46

It might help you to write \(t^2+1\) as yet another function \(y(t)\). Then you know that \(v(t)=w(y(t))\). So using the chain rule, you know that \(v'(t)=w\;'(y(t))y\;'(t)\)

Answer 47

i see it now

Answer 48

1/(x^3) flipped is 3/x?

Answer 49

For, 14, use the chain rule again. Since you also have the \(x^3\) function, you need to multiply by the derivative of \(x^3\).

Answer 50

i was using the rule: (ln(x))' = 1/x

Answer 51

First you need to use that, but you also need to use the chain rule like above. It also might help to rewrite the function as \(f(x)=g(h(x))\) where \(g(x)=\ln(x)\) and \(h(x)=x^3\). Then it should look very similar to number 13

Answer 52

can you plug it all in so i can see what you mean?

Answer 53

\[f'(x) = g'(h(x))h'(x) \Rightarrow f'(x) = {1\over {x^3}} \cdot 3x^2 ={3\over x}\]

Answer 54

ah i see

Answer 55

why is 16 E?

Answer 56

Because the very minimum of the function is at the point (2, 17). Since it's has an absolute value, that part of the function is always greater than or equal to 0. Then you have a +17, so the function is always greater than or equal to 17.

Answer 57

o ok. thats a good explanation.

Answer 58

why would 18 be C? i was thinking it would be D.

Answer 59

You can tell pretty quickly that it shouldn't be D or E because the exponent of \(e\) is positive. Since it's positive, \(5000e^{40}\) or \(5000e^{.40}\)are both greater than 5000. Since you have to gain money before you get to 5000, neither of those can be the correct answer.

Answer 60

im used to doing problems like this differently...
5000=P(e^8.0(5))

Answer 61

As for actually showing it's C, you want to solve for \(P_0\) in the following equation \[5000=P_0 e^{.08*5}=P_0e^{0.40}\]You can tell pretty easily that \[P_0 = ke^{-0.40}\]for some \(k\) since you need to get rid of the exponent. Thus, you know that \[5000=ke^{0.40} e^{-0.40}=ke^0=k\]Thus, \[P_0=5000e^{-0.40}\]

Answer 62

im a little confused by that explanation

Answer 63

Is there a specific part of it you're confused by?

Answer 64

You can tell pretty easily that
P0=ke−0.40
for some k since you need to get rid of the exponent. Thus, you know that
5000=ke0.40e−0.40=ke0=k
Thus,
P0=5000e−0.40

Answer 65

lol

Answer 66

So you can see why we're solving for \(P_0\) in \[5000=P_0 e^{0.40}\]then correct?

Answer 67

yes

Answer 68

since i cant give you any more medals here, i decided to give you some where you helped others. haha
but anyways, continue...

Answer 69

First thing to notice is that 5000 is an integer, while \(e^{0.40}\) is not. Thus, if you multiply \(P_0 e^{0.40}\) you have to get an integer. The best way to do this is let\[P_0=ke^x\]so that\[ke^x e^{0.40}=ke^{x+0.40}\]is an integer. By the properties of exponents, if you let \(x+0.40 =0\), then you have an integer. Solving for this, you get \(x=-0.40\)

Thanks for the medals :)

Answer 70

i still dont really understand this one but its ok...
can you explain 20 to me? i have no clue about that one.

Answer 71

I think it's easier than you're making it out to be. Since you have addition, it pretty straightforward. First, write out what \(h'(x)\) is. Namely, \[h'(x)=f'(x)+g'(x)\]Also, you're given above that the tangent lines at \(x=4\) are \[f'(4)=6x-4\]\[g'(4)=2x-2\]So the slopes of these at 6, and 2, respectively. Thus, \(6+2=8\) is \(h'(4)\)

Answer 72

you took the derivative of each and added them?

Answer 73

Basically. I probably shouldn't have used \(f'(4)\) and \(g'(4)\), but they shouldn't have used \(x\) in the equation either since it's a different variable.

But anyways, it says that the tangent line at \(x=4\) is given by those equations. You can easily tell that the slope of those lines is 6 and 2 by taking the derivatives. Since it's a straight line, you know that that is also the slope of \(f(x),\; g(x)\) at \(x=4\). Then you just need to add them together to solve for \(h'(4)\).

Answer 74

ok thanks

Answer 75

i think that is about it for now. thanks for being patient and helping me out!

Answer 76

You're welcome.