A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Need Help! Find derivative of f(x)=11/2sqrt(1x)
anonymous
 one year ago
Need Help! Find derivative of f(x)=11/2sqrt(1x)

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm really confused on how to find the derivative of fraction square root part..do I do the quotient rule?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0as much as i lectured before about not using rational exponents, that was when you had \[\frac{d}{dx}\sqrt{f(x)}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x) = 1\frac{ 1 }{ 2\sqrt{1x} }?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, that's it astrophysics!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for this one you probably do want to use rational exponents

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0LOL THANKS SATTELITE!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1\frac{1}{2}(1x)^{\frac{1}{2}}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1ye was gonna do that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess you are looking for the second derivative of the previous one right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow...so like why couldnt I use the rational exponents for the other question we were doing? and yea

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0of 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
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0this one would be a bear to do without the power rule, but let me lecture once again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439263414965:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Power rule and chain rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0DO NOT USE THE QUOTIENT RULE IF THE NUMERATOR IS A NUMBER !!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1^ this, I see so many people applying quotient rule to problems like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i will pontificate more on that in a moment, lets finish this one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay enough about bashing on mwah, let's finish the problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{4}(1x)^{\frac{3}{2}}\] is your derivative right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea and then I solved it down to what I have above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you trying to set it equal to zero?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well not you should not again the curse of rational exponents

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{4\sqrt{(1x)^2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{4\sqrt{(1x)^3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a fraction is only zero if the numerator is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0my guess is you are asked about concavity right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.0btw\[\left(\frac{1}{f}\right)'=\frac{f'}{f^2}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0oops \[\left(\frac{1}{x^2+2x}\right)'=\frac{2x+2}{(x^2+2x)^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay now im like super confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why cant I find its critical points, and find where f''(x) is undefined and thats it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0critical points are where the derivative is zero or undefined

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea so, hollup lemme draw it for ya

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to find where the function is increasing resp decreasing you need to know where the derivative is positive resp negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439264231886:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the problem here is the second derivative is NEVER 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439264312465:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lost 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
 one year ago
Best ResponseYou've already chosen the best response.0AND im right bubba because I just checked in the back of the book

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are you trying to find?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where the function is CU or CD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess I learned it a diff way than you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok good concave down for all x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well for all x in the domain i.e. for all x in \((\infty, 1)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0don't forget that 1 is the endpoint of the domain of the original function that means the derivative is undefined there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and since the derivative is undefined at \(x=1\) it would be an amazing miracle if the second derivative was defined there !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.