find the derivative of f(x)=x/sqrtx in 2 different ways
Hmm ok. So there are several ways to approach this. Have you tried anything yet? :o
One way is the quotient rule.
use the quotient rule and the product rule
basically the quotient rule can be derive using the product rule
derived*
\[\frac{ x }{ \sqrt x } = (\frac{ x }{ \sqrt x } ) \frac{ \sqrt x }{ \sqrt x } = \sqrt x\]
product rule's general form is f'(x) = u'v + uv' when applying a quotient into the product rule u/v you will end up with u * v^-1 u(x) * [v(x)]^-1
what on earth are you doing @jayz657
it can be simplified into sqrt(x) and you can just take the derivative of sqrt(x) using the power rule
True it does. I think he was commenting on the steps you showed. They're really strange :) lol
i multiplied by 1 since sqrt(x)/sqrt(x) = 1
you'll end up using the power rule when you derive the product rule
Oh ok, so I guess there was another step in there, cancelling the x's. I see :o
power rule and chain rule in fact
are you suggesting that f'(x) = sqrt (x) is the same as f'(x) = x/sqrt(x) ?
No. Jay was simplifying the problem before taking the derivative. He was asked to show several ways to take the derivative, so that seems like one nice approach.
yea i rationalized the denominator and it simplified to sqrt(x)
you're still stuck with deriving the same problem utilizing 2 different techniques here are the following: by the definition by product rule (needs chain and power rules) by quotient rule
\[\Large (\sqrt x)'\quad=\quad (x^{1/2})'\]Why would you need product or quotient rule to find that derivative? :\ Maybe I'm misunderstanding what you're saying.. You just use power rule for that one.
i would prob just use the quotient rule as the 2nd method
it is 1/sqrt(x) in this case which is why I said in my earlier post product rule's general form is f'(x) = u'v + uv' when applying a quotient into the product rule u/v you will end up with u * v^-1 u(x) * [v(x)]^-1
ya, 1/(2sqrt(x)) :o
I like how the asker hasn't replied at all LOL
either method works fine
XD
I know they all work fine the purpose of the exercise is to find out one's proficiency at using all derivative techniques
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