Question

h(u)=square (9-u^2)
I want to get the derivative.
Can't I take the 9-u^2 out of the square and make it 3-u?

Answer 1

o yes it can be written as (3-u)(3+u)

Answer 2

I thought so. Thanks

Answer 3

no... it isnt a square

Answer 4

(3-u) (3-u) does NOT equal (9-u^2) it equals (9 -6u +u^2)

Answer 5

try using trig substitutions...

sqrt(9 - u^2);

sqrt(9 - (3sin)(3sin)) ; u=3sin

sqrt(9 -9sin^2) = sqrt(9(1-sin^2)) = sqrt(9(cos^2)) = 3cos

Answer 6

the derivative of 3 cos is -3sin right? plug in your sin value now..

Answer 7

sin = u/3 ..thats what I meant :)

Answer 8

-3(u/3) = -u if I did it right....

Answer 9

or am i confusing integrals and derivatives.... one of these days Ill know when Im right :)

Answer 10

I get... -1/sqrt(9-u^2) is that better?

Answer 11

-u/sqrt(9-u^2) ??

Answer 12

(9-u^2) is not equals (9 -6u +u^2)
(9-u^2) is equal to (3-u )(3+u)

Answer 13

and the sqrt((3-u)(3+u)) is not (3-u)...

Answer 14

formula is a^2 - b^2 = (a+b)( a-b)

Answer 15

Is the sqrt(9-u^2) = (3-u)? No.

Answer 16

I might be wrong on everything else I typed, but I know im right on this one :)

Answer 17

the sqrt(9-u^2) is not equal to (3-u) but sqrt(9-u^2) is equal to (3-u)(3+u)

Answer 18

How so?

9-u^2 is equal to (3-u)(3+u)

sqrt(9-u^2) is NOT equal to (3-u)

square (3-u) and get (9-u^2) for me, just try it.... try it.... cant do it...

Answer 19

is [(3-u)(3+u)]^2 equal to..... where was I?

Answer 20

Couldn't you just square \[(9-u ^{2})^{2}\]

Answer 21

maybe..... just maybe...... are we allowed to? :)

Answer 22

[(3-u)(3+u)]^2 equal to (3-u)^2 (3+u)^2

Answer 23

Wouldn't that give you \[u ^{4}-18u ^{2}+81\]

Answer 24

Then do the derivative\[4u ^{3}-36u\]

Answer 25

.... im sure wali is right about something in this....

Answer 26

do we have to unsquare the derivative?

Answer 27

ok u win the game but i m not wrong

Answer 28

lol ... but I was gonna concede :)

Answer 29

maybe

Answer 30

"Can I take the 9-u^2 out of the square and make it 3-u?" What was the answer?

Answer 31

no no ...

Answer 32

I chose "no" as well, but im not a reliable witness :)

Answer 33

Then simplifying getting\[1/4u[(u-3)(u+3)]\]

Answer 34

ok by we will meet again because i have to prepare for tomorrow calculus paper by