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CPhill

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 #4
avatar+129899 
+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.....
5 апр. 2014 г.
 #1
avatar+129899 
+5
9+4*(8*8-30)
4 апр. 2014 г.
 #1
avatar+129899 
+2
1. Bakery sells tortillas and bagels and rolls. The ratio is 12: 5 :7. Number of tortillas sold is 50 more than rolls sold.
a. Find out how many tortillas bagels and rolls were sold altogether on that Saturday.

2. Weight of potatoes used by Mrs. Wilson is 5/2 times the weight of carrots used.
a. The weight of potatoes used was 9 pounds more than the weight of carrots used. Find the total weight of both ingredients.

3. A wall is painted yellow and brown. The area painted yellow is 3 times the area painted brown.
a. The wall has an area of 8 square meters. Find the area of the wall painted yellow.

4. Peter collects U.S. and foreign stamps. He has 5 times as many U.S. stamps as foreign stamps.
a. Peter has 140 more U.S. stamps than foreign stamps. How many stamps does he have in his collection?

I'll try to keep this as non-mathematical as possible.

1. Let's think about this one for a second. Note, that for every 12 tortillas sold, 7 rolls are sold. So, that's a difference of 5. So, to have a difference of 50, we'd have to sell 10 times as many of each. So, that must be 120 tortillas and 70 rolls. And, to find the number of bagels sold, we just note that that must be 5 times 10, too. So, we sold 120 tortillas, 50 bagels and 70 rolls.

2. This one is a little harder, but not too bad. Let's call the weight of carrots used "x." Then the weight of potatoes used = 5/2 * x. Then, since 9 more pounds of potatoes are used than carrots, we can express this difference in the following way:

(Weight of potatoes used) - (Weight of carrots used) = 9
Thus..........

(5/2 * x ) - (x) = 9
Then,
(3/2 * x) = 9
And multiplying both sides by 2/3, we have
x = 6
So, 6 lbs. of carrots were used. And (5/2 * x) = (5/2 * 6) = 15 lbs, of potatoes were used.

3. This one is easy. Call the area painted brown "x." The, the yellow painted area is 3x, and the total area painted is 8.
Thus......
(Area painted brown) + (Area painted yellow) = 8 square meters
x + 3x = 8
4x = 8
Dividing by 4 on each side, we get
x = 2......So 2 square meters are painted brown. And three times this much is painted yellow.....6 square meters..

4. This one is kinda' like number 2. Let's call the number of foreign stamps he has "x." Then, the number of US stamps he has is five times as many, or just 5x. And the difference between the two types is 140.

Then
(The number of US stamps) - (The number of foreign stamps) = 140.

Setting this up as an equation, we have
5x - x = 140
4x =140.
Divide both sides by 4
x = 35 ......This is the number of foreign stamps.
And 5x = the number of US stamps = 5(35) = 175.

Note that 175 - 35 = 140.
So the total number of stamps in his collection is just 175 + 35 = 210.

That wasn't too bad, huh???
31 мар. 2014 г.
 #3
avatar+129899 
+2
¡Para servirle!
30 мар. 2014 г.