Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

I am stuck with an easy problem - I must be missing something. I am trying to use the limit process to find the derivative of f(x) = 8 / sqrt of x. Can you help me see where I am messing up?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[f(x) = 8/\sqrt{x}\] find the derivative using the limit process
f'(x) = lim \[f^\prime(x) =\lim_{\Delta x \rightarrow 0} [(8/\sqrt{x+c}- 8/\sqrt{x})/c]\]
I am banging my head against this problem and getting a bad answer every time. We can tell by looking at it that the answer is\[-(4/x^(3/2))\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Well, I think you see what I mean...
But I haven't figured out the algebra to leave me with a 4 in the numerator. I keep getting stuck with things that don't work out in the end.
Weeku, what do you have for me so far?
Whoops, I made a mistake. Darn editor. . . Hold on.
thanks
Seems no matter what I try, I end up with delta x /[( sqrt x + delta x) + sqrt x]
I am going in circles.
are you still there?
Yeah. I'm stuck as well. Hrm. . . I'm still trying, though. Sorry xD Interesting problem, though.
Ah. Got it. Took a whole loose-leaf paper to work out. . . Pretty nasty.
Did you get up to \[(8\sqrt{x} - 8\sqrt{x+h}) / (h \sqrt{x}\sqrt{x+h})\]? Or, do you know how I got there? At that point, multiple the numerator and denominator by the conjugate of the numerator.
multiply*
Then it becomes a lot of distributing and factoring, until you can finally divide out the h. Once that happens, substitute 0 for h. Phew. I dunno if I can type out everything I did. . . I probably did it the long way, but I saw no other, shorter way. Do you want me to type it out?
Yes, I got that far. I see you have h = delta x and you canceled out one h already.
I would be grateful if you typed it out.
Before that, all I did was find the common denominator of 8/sqrt[x+h] and 8/sqrt[x], add them together, and instead of dividing by h, multiply by 1/h to get \[8\sqrt{x} - 8\sqrt{x+h} / h \sqrt{x} \sqrt{x+h}\] Then I factored out the 8 to get the above ^ Oh. Okay.
Whoops. Never mind about factoring out the 8.
I tried doing that, with no luck. :D
\[\frac{\frac{8}{\sqrt{x+h}} - \frac{8}{\sqrt{x}}}{h}\]Firs Only the Numerator.... \[8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\times \sqrt{x}}\] Rationalization of Numerator \[\frac{-8}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}\] \[\frac{-8}{2\times x^{\frac{3}{2}}}\]
Ishaan94, I am looking over what you have done to see where I diverged. Thank you.
I lost you after "Rationalization of Numerator," but nicely done. I did it the very, very long way, I see. . . Sorry about that, FreeTrader.
\[\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}\]
\[8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x} \times \sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}\]
I see how Ishaan94 got to the "first only the numerator..." expression. But I am lost on how he got rid of the terms in the numerator.
Right, multiply by conjugate.
I really think you are going about this the wrong way. You should change the square root into a power (1/2) and then bring the denominator to the top by changing the power to negative. Then you can simply use the limit process of: f(x)= 8x^(-1/2) f(x+h)=8(x+h)^(1/2) And plug this into the derivative formula. You should get rid of the terms in the denominator.
Then factor out a h and cancel.
\[8\frac{x - (x+h)}{\sqrt{x}\times \sqrt{x+h} \times (\sqrt{x} + \sqrt{x+h})}\]
I think it's dawning on me that I missed distributing the -1 on the h, too.
Stick with me, I am going to check that on my paper.
Ishaan94, when I perform the operation 8 *[ x -(x+h) ] I am left with 8x -8(x+h) which gives 8x -8x -8h
Right
Which gives -8h. divide out the h's.
I am onto this now... working the algebra. searching for the lightswitch for the bulb over my head...
\[f(x)=\frac{8}{\sqrt{x}}\] \[f(x)=\frac{8}{x ^{1/2}}\] \[f(x)=8x ^{\frac{-1}{2}}\] \[f(x+h)=8(x+h)^{-1/2}\] \[f \prime(x)\lim_{h \rightarrow 0}=I can't remember the formula\] Now just use you derivative formula, cancel where necisary, factor out a h, then cancel with the denominator.
Lol@chaise xD
Lol
My problem is now in the denominator of the numerator. I don't see a h to cancel the one against the -8.
You're forgetting that the entire equation is over h at this point. \[\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}\]
I have that now. Thanks. I am down to working with the remaining denom terms.
All terms on the right hand side of the minus sign in your derivative formula that have a x term should cancel with the terms. I think you're going about this the wrong way. Instead of having a quotient you should really bring the denominator to the top by making the power negative. Remember this rule? am^-x = a/m^x
Yes. I am already so deep in the prob that I am going to finish it with frax.
Here's what I have in the denominator, which I am tempted to misspell as demon-ator. \[\sqrt{x} * \sqrt{x} * \sqrt{x+h} + \sqrt{x} * \sqrt{x+h} * \sqrt{x+h} \]
Since the h values go to zero in the limit, the following is the denominator in the limit: \[2\sqrt{x}\sqrt{x}\sqrt{x}\]
\[f(x)=\frac{8}{\sqrt{x}}\] \[f(x)=\frac{8}{x ^{1/2}}\] \[f(x)=8x ^{\frac{-1}{2}}\] \[f(x+h)=8(x+h)^{-1/2}\] \[\lim_{h \rightarrow 0}f'(x)=\frac{f(x+h)-f(x)}{h}\] lim{h-->0) of f'(x) = (8(x+h)^(-1/2) - 8x^(-1/2))/h You can do the rest, I'm sure.
Yes, freetrader, (sqrt[x])^3 = x^(3/2)
Okay, I am losing my sanity. Can you help me remember if \[\sqrt{x^3} = \sqrt{x}\sqrt{x}\sqrt{x}\] ???
Yes - and this should be in the denominator of the equation once it has been solved. Notice how the ^(-1/2) becomes 1.5 when you increase the power by one. And yes it does FreeTrader
Yup. That is a true statement.
Stop writing it as a square root until you get to the final stage of the question. sqrt(x)^3 = x^(3/2)
I have temp insanity.
'Tis okay.:)
So, my answer took me 7 hours to reach.
It was totally worth it!
THANK YOU FRIENDS!
:)

Not the answer you are looking for?

Search for more explanations.

Ask your own question