Need Help! Find derivative of f(x)=1-1/2sqrt(1-x)

- anonymous

Need Help! Find derivative of f(x)=1-1/2sqrt(1-x)

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- schrodinger

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- anonymous

I'm really confused on how to find the derivative of fraction square root part..do I do the quotient rule?

- anonymous

oh no !!

- 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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## More answers

- Astrophysics

\[f(x) = 1-\frac{ 1 }{ 2\sqrt{1-x} }?\]

- anonymous

Yep, that's it astrophysics!

- anonymous

for this one you probably do want to use rational exponents

- anonymous

LOL THANKS SATTELITE!

- anonymous

\[1-\frac{1}{2}(1-x)^{-\frac{1}{2}}\]

- Astrophysics

ye was gonna do that

- anonymous

then go nuts

- anonymous

i guess you are looking for the second derivative of the previous one right?

- anonymous

wow...so like why couldnt I use the rational exponents for the other question we were doing? and yea

- 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

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

kk gotcha

- anonymous

this one would be a bear to do without the power rule, but let me lecture once again

- anonymous

|dw:1439263414965:dw|

- Astrophysics

Power rule and chain rule

- anonymous

DO NOT USE THE QUOTIENT RULE IF THE NUMERATOR IS A NUMBER !!

- Astrophysics

^ this, I see so many people applying quotient rule to problems like that

- anonymous

i will pontificate more on that in a moment, lets finish this one

- anonymous

Okay enough about bashing on mwah, let's finish the problem

- anonymous

\[-\frac{1}{4}(1-x)^{-\frac{3}{2}}\] is your derivative right?

- anonymous

Yea and then I solved it down to what I have above

- anonymous

are you trying to set it equal to zero?

- anonymous

yea...should I not

- anonymous

well not you should not
again the curse of rational exponents

- anonymous

\[-\frac{1}{4\sqrt{(1-x)^2}}\]

- anonymous

crap

- anonymous

\[-\frac{1}{4\sqrt{(1-x)^3}}\]

- anonymous

a fraction is only zero if the numerator is

- 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

my guess is you are asked about concavity right?

- anonymous

yea

- 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

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

oops \[\left(\frac{1}{x^2+2x}\right)'=-\frac{2x+2}{(x^2+2x)^2}\]

- anonymous

Okay now im like super confused

- anonymous

sorry
what point?

- anonymous

why cant I find its critical points, and find where f''(x) is undefined and thats it

- anonymous

critical points are where the derivative is zero or undefined

- anonymous

yea so, hollup lemme draw it for ya

- anonymous

to find where the function is increasing resp decreasing you need to know where the derivative is positive resp negative

- anonymous

|dw:1439264231886:dw|

- anonymous

the problem here is the second derivative is NEVER 0

- anonymous

|dw:1439264312465:dw|

- anonymous

and then x=1 :)

- 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

AND im right bubba because I just checked in the back of the book

- anonymous

???

- anonymous

what are you trying to find?

- anonymous

where the function is CU or CD

- anonymous

I guess I learned it a diff way than you

- anonymous

ok good
concave down for all x

- anonymous

well for all x in the domain i.e. for all x in \((-\infty, 1)\)

- anonymous

yep thats it!

- anonymous

don't forget that 1 is the endpoint of the domain of the original function
that means the derivative is undefined there

- anonymous

and since the derivative is undefined at \(x=1\) it would be an amazing miracle if the second derivative was defined there !

- 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

yep, well thanks!

- anonymous

yw

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