anonymous
  • anonymous
using the definition of a derivative find f(x) =-2/sqrt{x}
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[f(x)= -2/ \sqrt{x}\]
anonymous
  • anonymous
this will be a pain to write out but not hard to compute. you must use the definition yes?
anonymous
  • anonymous
i know that part i am having problems with rationalizing the denominator

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anonymous
  • anonymous
you need to compute \[lim_{h->0}\frac{\frac{-2}{\sqrt{x+h}}+\frac{2}{\sqrt{x}}}{h}\]
watchmath
  • watchmath
Good opportunity to practice your LaTeX satellite! :)
anonymous
  • anonymous
go ahead watchmath.
anonymous
  • anonymous
i am learning. eventually i will be fluent in it
anonymous
  • anonymous
i know that part the step after that
anonymous
  • 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
  • anonymous
just factor out by 2 thats all you had t say lol thanks so much
anonymous
  • anonymous
i knew that i was just doubting myself
anonymous
  • anonymous
i knew that i was just doubting myself
anonymous
  • anonymous
so we will just write \[lim_{h->0}\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\]
anonymous
  • anonymous
all divided by h of course. we worry about that last.
anonymous
  • anonymous
so now we rationalize
anonymous
  • anonymous
\[\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}\]
anonymous
  • anonymous
yes multiply numerator and denominator by the conjugate of th enumerator.
anonymous
  • anonymous
which of course is \[\sqrt{x}+\sqrt{x+h}\]
anonymous
  • anonymous
when you do that the numerator will just be \[x-x+h=h\]
anonymous
  • anonymous
and the denominator will be the product \[\sqrt{x} \sqrt{x+h}) (\sqrt{x}+\sqrt{x+h})\]
anonymous
  • anonymous
sorry am not repsonding i was working it out thanks i got it
anonymous
  • anonymous
thats ok i was typing. did you get the numerator correctly? it is just h
anonymous
  • anonymous
yes the x's cancel
anonymous
  • 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
  • anonymous
yes they 'cancel' meaning they add to zero. and of course dividing by h means the h goes in the denominator
anonymous
  • anonymous
the h cancels
anonymous
  • anonymous
so that last ugly thing i wrote is what you get when you rationalize the numerator. yes now the h cancels.
anonymous
  • anonymous
leaving \[\frac{1}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}\]
anonymous
  • anonymous
now you can replace h by 0 since you will not be dividing by 0.
anonymous
  • anonymous
to get... \[\frac{1}{(\sqrt{x}\sqrt{x})(\sqrt{x}+\sqrt{x})}\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[=\frac{1}{2x\sqrt{x}}\]
anonymous
  • anonymous
oh damn i made a mistake early on the numerator was \[x-(x+h)=-h\] not \[x-x+h\]
anonymous
  • anonymous
a very bush league mistake but easily rectified. just replace the 1 in the numerator by -1
anonymous
  • anonymous
sorry!
anonymous
  • anonymous
so the correct answer is \[\frac{-1}{2x\sqrt{x}}\]
anonymous
  • anonymous
then don't forget to multiply by the -2 at the end because we factored it out.
anonymous
  • anonymous
so the "final answer" as we say is \[\frac{1}{x\sqrt{x}}\] sorry it took a while
anonymous
  • anonymous
all steps clear?
anonymous
  • anonymous
sorry my computer is freezing idk if its thes site or what
anonymous
  • anonymous
sometime the site is funky. earlier today it certainly was.
anonymous
  • anonymous
if you have a question about any step post and i will respond
anonymous
  • anonymous
i understand thank you sooo much!

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