anonymous
  • anonymous
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?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
o yes it can be written as (3-u)(3+u)
anonymous
  • anonymous
I thought so. Thanks
amistre64
  • amistre64
no... it isnt a square

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amistre64
  • amistre64
(3-u) (3-u) does NOT equal (9-u^2) it equals (9 -6u +u^2)
amistre64
  • amistre64
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
amistre64
  • amistre64
the derivative of 3 cos is -3sin right? plug in your sin value now..
amistre64
  • amistre64
sin = u/3 ..thats what I meant :)
amistre64
  • amistre64
-3(u/3) = -u if I did it right....
amistre64
  • amistre64
or am i confusing integrals and derivatives.... one of these days Ill know when Im right :)
amistre64
  • amistre64
I get... -1/sqrt(9-u^2) is that better?
amistre64
  • amistre64
-u/sqrt(9-u^2) ??
anonymous
  • anonymous
(9-u^2) is not equals (9 -6u +u^2) (9-u^2) is equal to (3-u )(3+u)
amistre64
  • amistre64
and the sqrt((3-u)(3+u)) is not (3-u)...
anonymous
  • anonymous
formula is a^2 - b^2 = (a+b)( a-b)
amistre64
  • amistre64
Is the sqrt(9-u^2) = (3-u)? No.
amistre64
  • amistre64
I might be wrong on everything else I typed, but I know im right on this one :)
anonymous
  • anonymous
the sqrt(9-u^2) is not equal to (3-u) but sqrt(9-u^2) is equal to (3-u)(3+u)
amistre64
  • amistre64
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...
amistre64
  • amistre64
is [(3-u)(3+u)]^2 equal to..... where was I?
radar
  • radar
Couldn't you just square \[(9-u ^{2})^{2}\]
amistre64
  • amistre64
maybe..... just maybe...... are we allowed to? :)
anonymous
  • anonymous
[(3-u)(3+u)]^2 equal to (3-u)^2 (3+u)^2
radar
  • radar
Wouldn't that give you \[u ^{4}-18u ^{2}+81\]
radar
  • radar
Then do the derivative\[4u ^{3}-36u\]
amistre64
  • amistre64
.... im sure wali is right about something in this....
amistre64
  • amistre64
do we have to unsquare the derivative?
anonymous
  • anonymous
ok u win the game but i m not wrong
amistre64
  • amistre64
lol ... but I was gonna concede :)
amistre64
  • amistre64
maybe
amistre64
  • amistre64
"Can I take the 9-u^2 out of the square and make it 3-u?" What was the answer?
anonymous
  • anonymous
no no ...
amistre64
  • amistre64
I chose "no" as well, but im not a reliable witness :)
radar
  • radar
Then simplifying getting\[1/4u[(u-3)(u+3)]\]
anonymous
  • anonymous
ok by we will meet again because i have to prepare for tomorrow calculus paper by

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