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differentiate using the definition

Mathematics
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\[\frac{ 1 }{ \sqrt{x} }\]
it's lim as h->0 (f(x+h)-f(x))/h
f(x+h) = 1/sqrt(x+h) f(x) = 1/sqrt(x)

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Other answers:

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

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