anonymous
  • anonymous
Need Help! Find derivative of f(x)=1-1/2sqrt(1-x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I'm really confused on how to find the derivative of fraction square root part..do I do the quotient rule?
anonymous
  • anonymous
oh no !!
anonymous
  • anonymous
as much as i lectured before about not using rational exponents, that was when you had \[\frac{d}{dx}\sqrt{f(x)}\]

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Astrophysics
  • Astrophysics
\[f(x) = 1-\frac{ 1 }{ 2\sqrt{1-x} }?\]
anonymous
  • anonymous
Yep, that's it astrophysics!
anonymous
  • anonymous
for this one you probably do want to use rational exponents
anonymous
  • anonymous
LOL THANKS SATTELITE!
anonymous
  • anonymous
\[1-\frac{1}{2}(1-x)^{-\frac{1}{2}}\]
Astrophysics
  • Astrophysics
ye was gonna do that
anonymous
  • anonymous
then go nuts
anonymous
  • anonymous
i guess you are looking for the second derivative of the previous one right?
anonymous
  • anonymous
wow...so like why couldnt I use the rational exponents for the other question we were doing? and yea
anonymous
  • anonymous
of course you could, what i meant was it is not necessary once you know it because after you use the power rule, you are going to have to convert out of the rational exponents to do any sort of computation
anonymous
  • anonymous
i wasn't saying you couldn't use it, i was saying you SHOULDN'T use it because you should just memorize the derivative of the square root function or its composition
anonymous
  • anonymous
kk gotcha
anonymous
  • anonymous
this one would be a bear to do without the power rule, but let me lecture once again
anonymous
  • anonymous
|dw:1439263414965:dw|
Astrophysics
  • Astrophysics
Power rule and chain rule
anonymous
  • anonymous
DO NOT USE THE QUOTIENT RULE IF THE NUMERATOR IS A NUMBER !!
Astrophysics
  • Astrophysics
^ this, I see so many people applying quotient rule to problems like that
anonymous
  • anonymous
i will pontificate more on that in a moment, lets finish this one
anonymous
  • anonymous
Okay enough about bashing on mwah, let's finish the problem
anonymous
  • anonymous
\[-\frac{1}{4}(1-x)^{-\frac{3}{2}}\] is your derivative right?
anonymous
  • anonymous
Yea and then I solved it down to what I have above
anonymous
  • anonymous
are you trying to set it equal to zero?
anonymous
  • anonymous
yea...should I not
anonymous
  • anonymous
well not you should not again the curse of rational exponents
anonymous
  • anonymous
\[-\frac{1}{4\sqrt{(1-x)^2}}\]
anonymous
  • anonymous
crap
anonymous
  • anonymous
\[-\frac{1}{4\sqrt{(1-x)^3}}\]
anonymous
  • anonymous
a fraction is only zero if the numerator is
anonymous
  • anonymous
so forget about setting this equal to zero, it is never zero lesson: once you have the derivative (whatever method) get it out of exponential notation and write what it really is
anonymous
  • anonymous
my guess is you are asked about concavity right?
anonymous
  • anonymous
yea
anonymous
  • anonymous
ok the square root is always positive so minus the square root is always negative i.e. the second derivative is negative for all x in the domain
anonymous
  • anonymous
btw\[\left(\frac{1}{f}\right)'=-\frac{f'}{f^2}\] don't never ever use the quotient rule if the numerator is a constant\[\left(\frac{1}{x^2+2x}\right)'=-\frac{2x+2}{x^2+2x}\]
anonymous
  • anonymous
oops \[\left(\frac{1}{x^2+2x}\right)'=-\frac{2x+2}{(x^2+2x)^2}\]
anonymous
  • anonymous
Okay now im like super confused
anonymous
  • anonymous
sorry what point?
anonymous
  • anonymous
why cant I find its critical points, and find where f''(x) is undefined and thats it
anonymous
  • anonymous
critical points are where the derivative is zero or undefined
anonymous
  • anonymous
yea so, hollup lemme draw it for ya
anonymous
  • anonymous
to find where the function is increasing resp decreasing you need to know where the derivative is positive resp negative
anonymous
  • anonymous
|dw:1439264231886:dw|
anonymous
  • anonymous
the problem here is the second derivative is NEVER 0
anonymous
  • anonymous
|dw:1439264312465:dw|
anonymous
  • anonymous
and then x=1 :)
anonymous
  • anonymous
lost me on that one before we continue, let me point out that the function is always concave down that means the second derivative is always negative
anonymous
  • anonymous
AND im right bubba because I just checked in the back of the book
anonymous
  • anonymous
???
anonymous
  • anonymous
what are you trying to find?
anonymous
  • anonymous
where the function is CU or CD
anonymous
  • anonymous
I guess I learned it a diff way than you
anonymous
  • anonymous
ok good concave down for all x
anonymous
  • anonymous
well for all x in the domain i.e. for all x in \((-\infty, 1)\)
anonymous
  • anonymous
yep thats it!
anonymous
  • anonymous
don't forget that 1 is the endpoint of the domain of the original function that means the derivative is undefined there
anonymous
  • anonymous
and since the derivative is undefined at \(x=1\) it would be an amazing miracle if the second derivative was defined there !
anonymous
  • anonymous
what you are interested in is not where the second derivative is zero, but rather where it is positive or negative in this case it is never zero, it is negative for all x
anonymous
  • anonymous
yep, well thanks!
anonymous
  • anonymous
yw

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