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 #4
avatar+130561 
+7
1. y = (t^2 + 5)e^t find dy/dt (or y')
There are a couple of different ways we could do this one, but let's call the expression inside the parenthesis f and the other part g.
So, we have y = fg
And, using the product rule, we have y' = f 'g + f g '
so f ' = 2t and g ' = e^t
And putting this together gives us
y' = (2t)e^t + (t^2 + 5)e^t ..... And simplifying we have
y' = e^t(t^2 + 2t + 5)

2. R(x) = 11 - 2cos(Π x)
Find R ' (x)
OK......Let's just concentrate on the "cos(Π x) " part....we'll go back and finish the rest in a sec
Note, that we can use the chain rule here. Call cos(Π x) the "outside" function and call Π x the "inside" function,
So, the derivative of the "outside function" is just -sin(Π x) and the derivative of the "inside" function is just Π
Remembering that the derivative of the constant 11 is just zero, we have
R ' (x) = -2(-sin(Π x) * Π).....And simplifying this, we have
R ' (x) = 2Πsin(Π x)

3. w = sin(8e^t)
Find w'
This is pretty much like the last one. We're going to use the chain rule here, too. Call sin(8e^t) the "outside" function and call 8e^t the "inside" function.
Then the chain rule says to take the derivative of the "outside" times the derivative of the "inside."
So the derivative of sin(8e^t) is just cos(8e^t) and the derivative of 8e^t is just itself.
And putting this together we get
w' = cos(8e^t) * 8e^t or just
8e^t cos(8e^t)

4. f(x) = y = (2x + 1)^9 (3x - 1)^7 .....Find the first and second derivatives.
This one is more "tedious" than difficult. It combines both the chain and product rules (plus a little simplifying).
As before, let's call (2x + 1)^9, "f" and let's call (3x - 1)^7, "g."
Then, y = fg and y' = f 'g + f g '
So, to calculate f ' , we'll need to employ the chain rule. Let's call (2x + 1)^9 the "outside" function and 2x + 1 the "inside" function.
The derivative of the outside is 9(2x + 1)^8 and the derivative of the inside is just 2. And multiplying these together we get 18(2x + 1)^8.
Similarly, we'll need the chain rule again to find g'. Let (3x - 1)^7 be the "outside" function and let 3x - 1 be the "inside" function.
The derivative of the outside is 7(3x - 1)^6 and the derivative of the inside is just 3. And multiplying these together we get 21(3x - 1)^6.
So, putting all this together, we get.
y' = f 'g + f g '
y' = 18(2x + 1)^8 (3x - 1)^7 + (2x + 1)^9 (21(3x - 1)^6)
Let's factor out the lowest powers on the "(2x + 1)" and "(3x - 1)" terms.
This gives (2x + 1)^8 * (3x - 1) ^6 * [18(3x - 1) + 21(2x + 1)]
And simplifying inside the brackets we have (2x + 1)^8 * (3x - 1) ^6 * [96x + 3]
And factoring a "3" out of [96x + 3], we have
3*(2x + 1)^8 * (3x - 1) ^6 * (32x + 1)

To find y '', we just extend the product rule over one more function, i.e.
y' = fgh
y'' = f' gh + fg' h + fgh'
To make things easier, I'm going to put our last answer back into this form....... (2x + 1)^8 * (3x - 1) ^6 * [96x + 3], where "f" is (2x + 1)^8, "g" is (3x - 1) ^6, and "h" is [96x + 3]
To save some time, (And because I think you see how to apply the chain rule now), f ' = 8(2x + 1)^7 * (2) = 16(2x + 1)^7 and g ' = 6(3x - 1) ^5 *(3) = 18(3x - 1) ^5 and h' = 96
Sp...let's put everything together and simplify...we have
y'' = f' gh + fg' h + fgh'
y''= (16(2x + 1)^7) * (3x - 1) ^6 * [96x + 3] + (2x + 1)^8 * 18(3x - 1) ^5 * [96x + 3] + (2x + 1)^8 * (3x - 1) ^6 * [96]
So, (as before), factoring out the lowest powers of the (2x + 1 ) and (3x -1) terms, we have
(2x + 1)^7 * (3x - 1)^5 [(16 (3x - 1)* (96x + 3) +(18) (2x +1) (96x +3) + (2x + 1) (3x -1 ) (96)] =
(2x + 1)^7 * (3x - 1)^5 [ 16 (288x^2 - 87x - 3) + 18 (192x^2 + 102x + 3) + 96(6x^2+ x -1)]
(2x + 1)^7 * (3x - 1)^5 [8640x^2 + 540x - 90]
Finally, we can factor the last term as 10 * (864x^2 + 54X - 9) = 10 * 3 * ( 288x^2 + 18x - 3) = 10 * 3 * 3 * (96x^2 + 6x - 1) = 90 (96x^2 + 6x - 1)
So we get
90 *(2x + 1)^7 * (3x - 1)^5 * (96x^2 + 6x - 1)

Whew!!! That was a long one, huh??

I hope this helps.....
Apr 5, 2014
 #3
avatar+2354 
+5
Okay, so f(x) = (2x+1) 9(3x-1) 7.

Now we have two functions so we want to apply the product rule.

The product rule works for functions in the following way f(x) = g(x) * h(x)

in this case g(x) = (2x+1) 9 and h(x) = (3x-1) 7

Since we know the product rule states f'(x) = g'(x)*h(x) + g(x) * h'(x)

we have to calculate the derivatives of g(x) and h(x).

For both we need the chain rule which states f(x) = g(h(x)) => f'(x) = g'(h(x))*h'(x).

That equation might be an appaling sight so let me attempt to state it to you in simpler words.

The derivative of a function 2 within a function 1 is the derivative of the function 1 leaving the function 2 as it is multiplied by the derivative of the function 2

Since this still might seem a little fuzzy let me give you an example with the functions we have g(x) = (2x+1) 9

We could again rewrite this to g(x) = (u(x)) 9. where u(x) = 2x+1 (so u'(x) = 2) then g'(x) = 9*u(x) 8*u'(x) = 9*(2x+1) 8*2 = 18(2x+1) 8

Now, let me continue with the rest of the original question.

We just saw that g'(x) = 18(2x+1) 8

Similarly we can calculate (again using the chain rule) h'(x) = 7*(3x-1) 6*3 = 21*(3x-1) 6

Then for f'(x), we can fill in the blanks.

f'(x) = g'(x)h(x) + g(x)h'(x) = 18(2x+1) 8* (3x-1) 7 + (2x+1) 9*21(3x-1) 6

We can simplify this to 18(2x+1) 8*(3x-1) 6*(3x-1) + (2x+1) * (2x+1) 8 * 21(3x-1) 6

I did this so both terms have (2x+1) 8*(3x-1) 6 in common.

This means I can now simply the equation to (18(3x-1)+21(2x+1)) * (2x+1) 8*(3x-1) 6

Note that: 18(3x-1)+21(2x+1) = 54x-18+42x+21 = 96x+3 = 3(32x+1)

So we calculated that f'(x) = 3(32x+1)(2x+1) 8(3x-1) 6

f"(x) works in quite a similar way, where we can again use f(x) = g(x)h(x) only this time g(x) also is a multiplication of two functions so for g'(x) you'd also need to use the product rule.

Must be a case of productruleception

See whether you can do that one yourself.

Reinout

edit: just saw Alan did the same thing in the mean time
Apr 5, 2014
 #1
avatar+2354 
+5
Stu:

Online test questions, you don't have to give me the answer as I do want to learn the process and be able to do it, but I suspect that the way the test is set up, that investigation and getting help is part of the process to encourage learning of difficult material. So answers are not necessary but mighty helpful since, I can see what is wrong, thus feel free to put new variables in. Note there is a few questions here, challenging for me, and since I can not follow the derivative rules to well, or if I do the answers are not put it in the correct form etc. The last answers I got were input right so thank you and that shows that if the process is right the system will accept the answers.

1
http://snag.gy/w8UHZ.jpg

2
http://snag.gy/FrQ15.jpg

3
http://snag.gy/sCKWg.jpg

4
http://snag.gy/eE7hY.jpg
I really would like an explanation of this question and answer, or if there was a another answer to arrive at through a different method and what that method would be. thanks.

I appreciate the help.




Okay, let's see what I can do here,
I think I'll give you important hints for the first three and show you the whole solution to the fourth one to give you an idea.

Also I think it might help you to look at this site; http://www.mathsisfun.com/calculus/derivatives-rules.html
This website has lot's of simple explanations on various mathematical subjects and this section pretty much summarizes the rules for derivatives.



Now, for the first one, you should look into the product rule. The idea is that f(x) = g(x)*h(x) then f'(x) = g'(x)*h(x) + g(x)*h'(x)

So for example if g(x) = x 2 and h(x) = e x so f(x) = x 2*e x

then f'(x) = 2xe x + x 2e x = (2x+x 2)e x



For the second one, look at the derivatives of trigonometric functions (and learn them by heart). So, f(x) = sin(x) => f'(x) = cos(x) and g(x) = cos(x) => g'(x) = -sin(x)

Also have a look at the chain rule to see what to do with cos(pi*x).



For the third one,

Same as the second one, look at the derivatives of trigonometric functions and the chain rule. That should do the trick

I'll do the fourth one in a seperate post.
Apr 5, 2014
 #2
avatar+1313 
0
Doesn't this depend on what is the movable object and unstoppable force?
Apr 5, 2014
 #1
avatar+1313 
+3
bump
Apr 5, 2014
 #1
avatar+130561 
+5
9+4*(8*8-30)
Apr 5, 2014
Apr 4, 2014
 #33
avatar+2354 
+11
Here are the new probability puzzles!
--------------------------------------------------------------------------------------------------------------------------------------
Easier one:

Ben the hiker wants to spend the night on top of a mountain. He hikes all day long towards the top and camps there overnight. The following day he is taking the same path down towards the bottom when he suddenly looks at his watch en notices that he is at the very same spot at the very same time of the day as the day before on his way up. 'What a coïncidence!' he exclaims!
Give an estimate of the probabilty that a hiker will be at exactly the same spot on the mountain at the same time of the day on his return trip as he was the previous day's hike up the mountain and explain why.

--------------------------------------------------------------------------------------------------------------------------------------
Hard one :

You and your best friend have been arrested for several white collar crimes. (Or something else which is more classy than stealing a candy bar)
Your attorney has told you the police has only got enough evidence to put you and your friend behind bars for 3 years.
However, the both of you are given the option to confess. If both of you confess you will both serve 10 years,
if one of you confesses and the other does not, the one who did not confess will serve 20 years.
Obviously, if you both remain silent you will both serve 3 years.
You are not allowed to have any communication with your friend.
Suppose your main priority is to keep yourself out of jail, and your secondary priority is to keep your partner out of jail.
You have decided that you find it twice as important to keep yourself out of jail as you find it important to keep your partner out of jail.
(So for example, in your opinion him serving 10 years is equal to you serving 5).
You estimate the chance of your partner confessing p.
For which values of p will you choose to confess and for which values will you keep silent?

--------------------------------------------------------------------------------------------------------------------------------------

Reinout-g

edit; I changed 'at what value of p' to 'for which values of p'
Apr 4, 2014
 #92
avatar+2354 
+6
Melody:

Hi everyone,
My last wrap was 3 days ago. I was not really intending to do one tonight. The regular posts were about to 'fall off the page' and I was seriouously considering leaving them there.
Anonymous4338 made me re-think. Thank you for your post on this thread. I am glad I made you happy. You have made me happy too.

Since my last wrap many great answers have been posted by Samgelinas, CPhill, Alan, Kitty<3, reinout-g, brabrams, guest, Rom and Anonymous4338. Thank you.

Samgelinas posted an addictive web game. Thanks for that. I have added the thread address to my collective 'games' thread.

I have written a post 'Information for new members' on the 'Become a member' thread. I have talked about a few things there, I can't actually remember what - You will have to take a look.
I have however explained how the private message system works. Unless you have any particular reason not to, it would be a good idea if you set an alert pop-up for new messages.
I have explained it all in the post. I sometimes want to contact members privately. This is the only way I can do it. Other people may also want to chat.
http://web2.0calc.com/questions/become-a-member-why-would-you-want-to
I think that is it for now.
Stay happy,
Melody.



Thanks Melody

I like the end of the day wrap up because it is usually there when I open the forum (which is usually around 2pm.)

It saves me the time of skimming through all posts which happened during the night (such as someone who gave an interesting theory or answer and also might remind me to eg. update the probability puzzles (which I totally forgot) or have another look at the calculator options, etc. In addition to that it gives me an idea of when you're about the log-out, which is usually when I refresh the forum a little more often.

Moreover when I first came to the forum it also motivated me to answer more questions since it kind off felt like an achievement to be in the wrap up.

Nevertheless I do understand that the forum has become less active and that it might be too much to post one every day.
Perhaps you could post one every once in a while just for the update.

Reinout
Apr 4, 2014

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