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?

- anonymous

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- anonymous

\[f(x) = 8/\sqrt{x}\]
find the derivative using the limit process

- anonymous

f'(x) = lim \[f^\prime(x) =\lim_{\Delta x \rightarrow 0} [(8/\sqrt{x+c}- 8/\sqrt{x})/c]\]

- anonymous

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\[-(4/x^(3/2))\]

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## More answers

- anonymous

Well, I think you see what I mean...

- anonymous

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.

- anonymous

Weeku, what do you have for me so far?

- anonymous

Whoops, I made a mistake. Darn editor. . . Hold on.

- anonymous

thanks

- anonymous

Seems no matter what I try, I end up with delta x /[( sqrt x + delta x) + sqrt x]

- anonymous

I am going in circles.

- anonymous

are you still there?

- anonymous

Yeah. I'm stuck as well. Hrm. . . I'm still trying, though. Sorry xD Interesting problem, though.

- anonymous

Ah. Got it. Took a whole loose-leaf paper to work out. . . Pretty nasty.

- anonymous

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.

- anonymous

multiply*

- anonymous

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?

- anonymous

Yes, I got that far.
I see you have h = delta x and you canceled out one h already.

- anonymous

I would be grateful if you typed it out.

- anonymous

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.

- anonymous

Whoops. Never mind about factoring out the 8.

- anonymous

I tried doing that, with no luck. :D

- anonymous

\[\frac{\frac{8}{\sqrt{x+h}} - \frac{8}{\sqrt{x}}}{h}\]Firs Only the Numerator....
\[8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\times \sqrt{x}}\] Rationalization of Numerator
\[\frac{-8}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}\]
\[\frac{-8}{2\times x^{\frac{3}{2}}}\]

- anonymous

Ishaan94, I am looking over what you have done to see where I diverged. Thank you.

- anonymous

I lost you after "Rationalization of Numerator," but nicely done. I did it the very, very long way, I see. . . Sorry about that, FreeTrader.

- anonymous

\[\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}\]

- anonymous

\[8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x} \times \sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}\]

- anonymous

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.

- anonymous

Right, multiply by conjugate.

- chaise

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.

- chaise

Then factor out a h and cancel.

- anonymous

\[8\frac{x - (x+h)}{\sqrt{x}\times \sqrt{x+h} \times (\sqrt{x} + \sqrt{x+h})}\]

- anonymous

I think it's dawning on me that I missed distributing the -1 on the h, too.

- anonymous

Stick with me, I am going to check that on my paper.

- anonymous

Ishaan94, when I perform the operation 8 *[ x -(x+h) ]
I am left with 8x -8(x+h)
which gives 8x -8x -8h

- anonymous

Right

- anonymous

Which gives -8h. divide out the h's.

- anonymous

I am onto this now... working the algebra. searching for the lightswitch for the bulb over my head...

- chaise

\[f(x)=\frac{8}{\sqrt{x}}\]
\[f(x)=\frac{8}{x ^{1/2}}\]
\[f(x)=8x ^{\frac{-1}{2}}\]
\[f(x+h)=8(x+h)^{-1/2}\]
\[f \prime(x)\lim_{h \rightarrow 0}=I can't remember the formula\]
Now just use you derivative formula, cancel where necisary, factor out a h, then cancel with the denominator.

- anonymous

Lol@chaise xD

- anonymous

Lol

- anonymous

My problem is now in the denominator of the numerator.
I don't see a h to cancel the one against the -8.

- anonymous

You're forgetting that the entire equation is over h at this point.
\[\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}\]

- anonymous

I have that now. Thanks. I am down to working with the remaining denom terms.

- chaise

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

- anonymous

Yes. I am already so deep in the prob that I am going to finish it with frax.

- anonymous

Here's what I have in the denominator, which I am tempted to misspell as demon-ator.
\[\sqrt{x} * \sqrt{x} * \sqrt{x+h} + \sqrt{x} * \sqrt{x+h} * \sqrt{x+h} \]

- anonymous

Since the h values go to zero in the limit, the following is the denominator in the limit:
\[2\sqrt{x}\sqrt{x}\sqrt{x}\]

- chaise

\[f(x)=\frac{8}{\sqrt{x}}\]
\[f(x)=\frac{8}{x ^{1/2}}\]
\[f(x)=8x ^{\frac{-1}{2}}\]
\[f(x+h)=8(x+h)^{-1/2}\]
\[\lim_{h \rightarrow 0}f'(x)=\frac{f(x+h)-f(x)}{h}\]
lim{h-->0) of f'(x) = (8(x+h)^(-1/2) - 8x^(-1/2))/h
You can do the rest, I'm sure.

- anonymous

Yes, freetrader, (sqrt[x])^3 = x^(3/2)

- anonymous

Okay, I am losing my sanity. Can you help me remember if \[\sqrt{x^3} = \sqrt{x}\sqrt{x}\sqrt{x}\] ???

- chaise

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

- anonymous

Yup. That is a true statement.

- chaise

Stop writing it as a square root until you get to the final stage of the question.
sqrt(x)^3 = x^(3/2)

- anonymous

I have temp insanity.

- anonymous

'Tis okay.:)

- anonymous

So, my answer took me 7 hours to reach.

- anonymous

It was totally worth it!

- anonymous

THANK YOU FRIENDS!

- anonymous

:)

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