find the derivative of y = sqrt(3)+(x)^1/3 + 1/x???

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find the derivative of y = sqrt(3)+(x)^1/3 + 1/x???

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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 \[\sqrt{3}\]?

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Other answers:

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
oh i did not give a complete answer. sorry. i only said derivative of \[\sqrt{3}\] is 0 the others need power rule.
(1/3)x^-2/3-1x^-2
however, \[\frac{1}{x}\] os a very common function, so you should just remember its derivative without using power rule. the derivative of \[\frac{1}{x}\] is \[-\frac{1}{x^2}\]
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
comes up so often you do not want to compute it afresh each time. for \[x^{\frac{1}{3}}\] youi use the power rule to get \[\frac{1}{3}x^{-\frac{2}{3}}\]
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!!!!!!!!!!
lets do them one at a time. you want to use the power rule, which says that the derivative of \[x^r\] is \[r\times x^{r-1}\] ok so far?
wat is de power rule??? haven't heard bout it
it tells you how to find the derivative of something raised to a power. for example the derivative of \[x^3\] is \[3x^2\]
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???
this one is the confusing one, but don't be fooled. \[\sqrt{3x}=\sqrt{3} \times \sqrt{x} = \sqrt{3}\times {x^{\frac{1}{2}}}\] and a constant just stays there.
k got it till here
for example, the derivative of \[x^2 \] is \[2x\] and the derivative of \[\sqrt{3}x^2\] is \[2 \sqrt{3}x\] the constant just says as a multiplier, so ignore it.
k so i ignore sqrt(3)
so only use the power rule on the \[x^\frac{1}{2}\] part. bring out the exponent as a multiplier, and then subtract 1 from the exponent. i wait while you try it.
but now did u take \[\sqrt{3}x ^{2}\] come into picture
u took it as an eg???
that was just an example to explain that the \[\sqrt{3}\] is unimportant. just a side example. not part of this problem.
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
ok. i explain. the exponent has a minus sign, so that means take the reciprocal. for example, \[x^{-5}=\frac{1}{x^5}\]
yups got it
the exponent is a fraction. the denominator is 2, so that means take the square root. the numerator is 1, so raise it to the power of 1, which is like doing nothing. so \[x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}\]
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!!!!!!
just \[\sqrt{x}\] in the denominator. but also a 2 in the denominator because you are multiplying by 1/2
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
ok. the reason this problem is a pain for you is that you have to do three things: 1)convert to exponential form 2) use the power rule 3) convert back to radical form
if u can hang around then give me some time i will just go through dese rules!!!!!!!!
but you probably know what \[7\times 8\] is because you have it memorized. since you are taking calculus, and since \[\sqrt{x}\] is such a common function, you should probably memorize its derivative, which is \[\frac{1}{2\sqrt{x}}\] that way you never have to do this again!
saves you the three steps of writing in exponential form, using the power rule, and converting back. if you remember it then for homework or on a test you just write it.
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!!!!!!!!
note the "-" sign. you can do this using the power rule as well, but it never changes. \[\frac{1}{x}=x^{-1}\] power rule gives \[-1\times x^{-2}=-\frac{1}{x^2}\]
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!!!!!!!
probably in the morning if you are here then. review power rule and of course exponents (because that is what it uses) in the mean time. good luck!
no i am in us.
k so wat's de time dere now????????
thanx for ur help and the pains u took but if u cud come online at the same time as u came 2day den it wud be gr8
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
1 Attachment
avoid the torn part look at the 1 written below!!!!!!!
looks good to me. of course you still have actually subtract the exponents and convert back to radical form.
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
and first one: \[\sqrt{3}\frac{1}{2}x^{\frac{1}{2}-1}=\sqrt{3}\frac{1}{2}x^{-\frac{1}{2}}=\frac{\sqrt{3}}{2\sqrt{x}}\]
ok?
yups
now try and easy one for yourself.
dude can u pls give me ur email id so dat de next time i have a doubt i can send it across or even chat again DUDE U ROCK!!!!!!!!! THANX FOR DE PATIENCE, SUPPORT.
no problem. you can email me here. best bet to catch me.
hold on i send an email address.
If u don't mind can u even give me ur landline or phone no my parents won't mind if i called some1 who can really help me through sticky situations!!!!!!!! and by the way a must watch!!!!! http://www.youtube.com/watch?v=kVFdAJRVm94
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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

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