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\[\sqrt{x}+x^{\frac{1}{3}}+\frac{1}{x}\]?

so the problem is
A) \[y=\sqrt{3+x ^{1/3}+1/x}\]
or
B)
\[y=\sqrt{3}+x ^{1/3}+1/x\]

or is first term \[\sqrt{3}\]?

the b) style

then do not be hoodwinked. \[\sqrt{3}\] is constant, its derivative is 0

yeap derivative of a constant is zero

satelllite ur dis ans is wrong it does not tally wid de book pls explain in detail

(1/3)x^-2/3-1x^-2

pls don't be angry at me cause i am in 11th and just a learner!!!!!!

Guys I don't know how to write the equation here coz I am bad at latex stuffs

k i will telll wat ans de book has

I hope tejeshwar95 got the answer

it says -
1/2*sqrt(3/x)+1/3*1/x^(2/3) - 1/x^2

which is the same as \[\frac{1}{2 \sqrt [3]{x^2}}\]

can't understand a thing

sorry lets go slow. is the first term just \[\sqrt{3}\]
or is it \[\sqrt{3x}\]

sorry for such a rude reply but i really can't get a hang of it!!!!!
its sqrt(3x)

ooooh then i apologize. i thought it was just \[\sqrt{3}\]

its ok!!!!!!!!!!

wat is de power rule??? haven't heard bout it

ok?

ok the 1 which says if y = x^n then n(x)^n-1

yups got it till here!!!!!!

yes that is the power rule.

so the trick is to write each of these terms in terms of exponents and then use the power rule.

k let me try it on paper hang on for a minute!!!!!!

now \[x^{\frac{1}{3}}\] is already written that way so that one is easy.

i will wait.

as far as \[\sqrt{3}\] isconcerned we remove the 3??? cause we can't differentiate it???

k got it till here

k so i ignore sqrt(3)

but now did u take \[\sqrt{3}x ^{2}\] come into picture

u took it as an eg???

yes just eg.

k
now after trying i got 1/2*x ^-1/2

yes!

now convert back to radical from from exponential form.

k correct till here sorry for being so slow

no problems. long as you understand.

do the same again??????

do you know what \[x^{-\frac{1}{2}}\] is in radical form?

if not i explain, if so just convert back.

nope not exactly

yups got it

therefore \[\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\]

so far so good?

so we put \[\sqrt{x^3}\] or \[\sqrt{x}\]

in the denominator!!!!!!

and the numerator is this case is \[\sqrt{3}\] because that constant is still there.

can we have a voice chat????
dis is becoming a pain!!!!!!

final answer:
\[\frac{\sqrt{3}}{2\sqrt{x}}\]

don't know how to voice chat. do you?

nope

voltage drops are happening i might not reply in between

if u can hang around then give me some time i will just go through dese rules!!!!!!!!

k learnt

and derived

and another very common function is \[f(x)=\frac{1}{x}\]

its derivative is \[f'(x)=-\frac{1}{x^2}\]

k got it

but could not derive it!!!!!!!!

when do u come online tomorrow i am tired i need to relax!!!!!!!!!!

or give me ur no if u live in delhi
den i can talk 2 u over de phone!!!!!!!

no i am in us.

k so wat's de time dere now????????

ok i will try. look for me around this time or a little earlier.

like wat's de time in US?? then i can guess wen 2 come online!!!!!!

or can u hang around for a while ny the time i play a bit of COD

k luks like i got it!!!!!!!!!!!

ok i will be here for a while. have some work to do.

i got the simplified form but ain't getin de full ans!!!!!!!!!!!

to part 1?

I got till see attachment

avoid the torn part look at the 1 written below!!!!!!!

how do i do dat pls tell!!!!!!

lets do the middle one.
\[\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{-\frac{2}{3}}\]

so far so good?

\[\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^2}}\]

yups

last one:
\[-1x^{-1-1}=-1x^{-2}=\frac{-1}{x^2}\]

yups

luks like i got the whole answer

ok?

yups

now try and easy one for yourself.

no problem. you can email me here. best bet to catch me.

hold on i send an email address.

Dude u online???? i hope i didn't harm ur feelings!!!!!!!!

no no i am still here. you can email at
satellite73.openstudy@gmail.com