anonymous
  • anonymous
differentiate using the definition
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ 1 }{ \sqrt{x} }\]
bahrom7893
  • bahrom7893
it's lim as h->0 (f(x+h)-f(x))/h
bahrom7893
  • bahrom7893
f(x+h) = 1/sqrt(x+h) f(x) = 1/sqrt(x)

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bahrom7893
  • bahrom7893
Do some algebra, cancel out the hs
anonymous
  • anonymous
\[\frac{ \frac{ 1 }{ \sqrt{x+h} }-\frac{ 1 }{ \sqrt{x} } }{ h }\]
bahrom7893
  • bahrom7893
yea, now multiply by the conjugate i think.. both top and bottom
anonymous
  • anonymous
\[\frac{ \frac{ \sqrt{x}+\sqrt{x-h}}{ \sqrt{x}\sqrt{x+h} } }{ h }\]
across
  • across
Typo up there ^
anonymous
  • anonymous
oops meant to put the h on the bottom :}
bahrom7893
  • bahrom7893
it's supposed to be a minus
bahrom7893
  • bahrom7893
all the way on top in the middle
anonymous
  • anonymous
oh ya that to
bahrom7893
  • bahrom7893
and the other one on top must be a plus
across
  • across
Inside the radical, too.
bahrom7893
  • bahrom7893
u mixed up the signs
anonymous
  • anonymous
oops!
anonymous
  • anonymous
ok so i got all that part, the conjugates are what get me
bahrom7893
  • bahrom7893
twiddla time then.. hang on a sec
anonymous
  • anonymous
\[\frac{ -h }{ h \sqrt{x}\sqrt{x+h(\sqrt{x}+\sqrt{x+h})} }\]
anonymous
  • anonymous
h goes away
anonymous
  • anonymous
that square root is mess up in the bottom
bahrom7893
  • bahrom7893
yup
bahrom7893
  • bahrom7893
No, since h is 0, on the bottom u just have sqrt(x+0)
bahrom7893
  • bahrom7893
so it's -1/sqrt(x) which is correct
anonymous
  • anonymous
at what point do i make h = 0?
bahrom7893
  • bahrom7893
when u dont run into trouble if u do.
bahrom7893
  • bahrom7893
1/h <- NO 1/(4+1) <- YES

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