using the definition of a derivative find f(x)
=-2/sqrt{x}

- anonymous

using the definition of a derivative find f(x)
=-2/sqrt{x}

- chestercat

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- anonymous

\[f(x)= -2/ \sqrt{x}\]

- anonymous

this will be a pain to write out but not hard to compute. you must use the definition yes?

- anonymous

i know that part i am having problems with rationalizing the denominator

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## More answers

- anonymous

you need to compute
\[lim_{h->0}\frac{\frac{-2}{\sqrt{x+h}}+\frac{2}{\sqrt{x}}}{h}\]

- watchmath

Good opportunity to practice your LaTeX satellite! :)

- anonymous

go ahead watchmath.

- anonymous

i am learning. eventually i will be fluent in it

- anonymous

i know that part the step after that

- anonymous

ok lets give me a break and at least factor out the -2, since this has nothing to do with the limit ok?

- anonymous

just factor out by 2 thats all you had t say lol thanks so much

- anonymous

i knew that i was just doubting myself

- anonymous

i knew that i was just doubting myself

- anonymous

so we will just write \[lim_{h->0}\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\]

- anonymous

all divided by h of course. we worry about that last.

- anonymous

so now we rationalize

- anonymous

\[\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}\]

- anonymous

yes multiply numerator and denominator by the conjugate of th enumerator.

- anonymous

which of course is \[\sqrt{x}+\sqrt{x+h}\]

- anonymous

when you do that the numerator will just be
\[x-x+h=h\]

- anonymous

and the denominator will be the product
\[\sqrt{x} \sqrt{x+h}) (\sqrt{x}+\sqrt{x+h})\]

- anonymous

sorry am not repsonding i was working it out thanks i got it

- anonymous

thats ok i was typing. did you get the numerator correctly? it is just h

- anonymous

yes the x's cancel

- anonymous

the denominator is that ugly thing i wrote last. so you have in total
\[\frac{h}{h(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h}}\]

- anonymous

yes they 'cancel' meaning they add to zero. and of course dividing by h means the h goes in the denominator

- anonymous

the h cancels

- anonymous

so that last ugly thing i wrote is what you get when you rationalize the numerator. yes now the h cancels.

- anonymous

leaving
\[\frac{1}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}\]

- anonymous

now you can replace h by 0 since you will not be dividing by 0.

- anonymous

to get...
\[\frac{1}{(\sqrt{x}\sqrt{x})(\sqrt{x}+\sqrt{x})}\]

- anonymous

yes

- anonymous

\[=\frac{1}{2x\sqrt{x}}\]

- anonymous

oh damn i made a mistake early on
the numerator was \[x-(x+h)=-h\]
not \[x-x+h\]

- anonymous

a very bush league mistake but easily rectified. just replace the 1 in the numerator by -1

- anonymous

sorry!

- anonymous

so the correct answer is
\[\frac{-1}{2x\sqrt{x}}\]

- anonymous

then don't forget to multiply by the -2 at the end because we factored it out.

- anonymous

so the "final answer" as we say is
\[\frac{1}{x\sqrt{x}}\] sorry it took a while

- anonymous

all steps clear?

- anonymous

sorry my computer is freezing idk if its thes site or what

- anonymous

sometime the site is funky. earlier today it certainly was.

- anonymous

if you have a question about any step post and i will respond

- anonymous

i understand thank you sooo much!

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