Question

Easy derivitives question PLEASE HELP NEED TO CHECK MY ANSWER ON #7

https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxtYmFybmVzbWF0aHxneDoxN2Q0MzMwNTMxZjY3MThj

Answer 1

follow the link I have a question about #7 part A. Would the radius be Radical3?

Answer 2

@zepdrix please help im sure it would only take a second and im in dire need. This determines whether i get a C or a D in AP calculus

Answer 3

The radius for #7 should be 2*radical 3.
Take the derivative of both sides wrt time. Find the rate of change of the area.

Answer 4

how did you find that? @iPwnBunnies

Answer 5

6pi = 1/2*pi*r^2

6 = 1/2*r^2

12 = r^2

Square root both sides. sqrt(12) = 2*sqrt(3)

Answer 6

you dont have to use DA/DT*6pie=.5pier^2Dr/dt

Answer 7

A = 1/2 pi r^2

A' = 2/2 pi r r' they give you r' and the needed information for r

Answer 8

got it thanks. Can someone help me with 5 by any chabce part A?

Answer 9

product rule ... whats the definition?

Answer 10

haha I like the JYA notation, I've never seen that before.

Answer 11

yeah but its a division rule within a product rule no?

Answer 12

figure out the main process first, then fill in the smaller parts

Answer 13

h = fg

what is h' ?

Answer 14

you have to find the product rule between f(x) and g(x) and then derivite that once to get h'?

Answer 15

well, we can define h by 3 products, the quotient rule is over rated ...

h = x^(-1) (x^2 + 1) (4x+7)

Answer 16

Lies :O
The quotient rule is so awesome.

Answer 17

im thinking of a word that starts with an s, and rhyms with ducks

Answer 18

but if you bring it to the top wouldnt you be addint it not multiplying it?

Answer 19

1/x = x^(-1) nt sure what you mean by adding it

Answer 20

if we bring it to the top, then its just a product of 3

Answer 21

h = x^(-1) (x^2 + 1) (4x+7)

h' = [x^(-1)]' (x^2 + 1) (4x+7)
+ x^(-1) [(x^2 + 1)]' (4x+7)
+ x^(-1) (x^2 + 1) [(4x+7)]'

Answer 22

ah ok. and just wanna make sure the answer to 7 part a was 2rad3

Answer 23

2 sqrt3 was the radius value ...

Answer 24

A' = pi r r'

Answer 25

thanks i got .5pierad3

Answer 26

seems fair 2sqrt3 pi/4 = sqrt3 pi/2

Answer 27

thanks. And im not too sure how you got the x to go on top. Can you explain that a little more?for 5A

Answer 28

f = (x^2+1)/x but 1/x = x^(-1)
= x^(-1) (x^2+1)

Answer 29

oh ok. thanks.

Answer 30

yep

Answer 31

how do you multiply all of them out?

Answer 32

@amistre64 so is the first term -x^0-1x^-2(4x+7)

Answer 33

h = x^(-1) (x^2 + 1) (4x+7)

h' = [x^(-1)]' (x^2 + 1) (4x+7)
+ x^(-1) [(x^2 + 1)]' (4x+7)
+ x^(-1) (x^2 + 1) [(4x+7)]'


h' = -x^(-2) (x^2 + 1) (4x+7)
+ x^(-1) (2x) (4x+7)
+ x^(-1) (x^2 + 1) (4)

simplify to your hearts desire