Question

So no one out there doesnt know how to do my problem

Answer 1

nope, not unless you tell us what your problem is.

Answer 2

taking the derivative of x^(2/3) using lim f(x+h)-f(x)/h

Answer 3

Seriously, you can't do it by hand.

Answer 4

No

Answer 5

f(x+h) = (x+h)^2/3
f(x+h)-f(x) = (x+h)^2/3 - x^2/3

Answer 6

wow site is going nuts

Answer 7

let me get another browser and then work it out

Answer 8

ok

Answer 9

idea is to rationalize the numerator by multiplying top and bottom by
\[\sqrt[3]{x+h)^2}+\sqrt[3]{x^2}\]

Answer 10

wow maybe it is better now. having a fit

Answer 11

that wouldn't work in this case. its cube root. if only it were square root.

Answer 12

oh right sorry. hold on

Answer 13

difference of two cubes i guess is what you have to do

Answer 14

see this page for rationalizing numerators/denominators that have sums/differences of cube roots

http://www.helpalgebra.com/articles/rationalizedenominator.htm

scroll down to Example 11

Answer 15

so what will the result look like?

Answer 16

yeah it is the difference of two cubes you have to use. you use
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 17

in this case you have
\[a-b\] which is
\[(x+h)^{\frac{2}{3}}-x^{\frac{2}{3}}\] and you want
\[a^3-b^3\]which will be
\[(x+h)^2-x^2\]

Answer 18

(x+h)^2-x^2??

Answer 19

so in the right hand side of the identity for the difference of two cubes put
\[a=(x+h)^{\frac{2}{3}}\]
\[b=x^{\frac{2}{3}}\]

Answer 20

ok lets go slow

Answer 21

first of all the answer is very simple. the answer is
\[\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{\sqrt[3]{x}}\]

Answer 22

now the question is how to get there from the definition right?

Answer 23

yes

Answer 24

so somehow you have to get rid of those annoying cube roots in the numerator yes?

Answer 25

yes

Answer 26

ok so if you had
\[\sqrt{x+h}-\sqrt{x}\] you would multiply by the conjugate
\[\sqrt{x+h}+\sqrt{x}\] and why does this work? it works because
\[(a-b)(a+b)=a^2-b^2\] so if you have square roots now they will be gone

Answer 27

but you do not have square roots. you have cubed roots!

Answer 28

yep

Answer 29

so this method will not work and you cannot use the formula for the difference of two squares, you have to use the formula for the difference of two CUBES

Answer 30

ok

Answer 31

you want to make it look like
\[a^3-b^3\] because that way the cube roots will be gone

Answer 32

in other words you want to get
\[((x+h)^{\frac{2}{3}})^3-(x^{\frac{2}{3}})^3\]
which will give you
\[(x+h)^2-x^2\] no more radical

Answer 33

oh, ok

Answer 34

so you have to use
\[(a-b)(a^2+ab+b^2)\] and that will give it to you
that is, if you multiply top and bottom by
\[a^2+ab+b^2\] you will get what you want

Answer 35

in this case we have
\[a=(x+h)^{\frac{2}{3}}\]
\[b=x^{\frac{2}{3}}\]
\[a-b=(x+h)^{\frac{2}{3}}-x^{\frac{2}{3}}\]

Answer 36

so now slowly

\[a=(x+h)^{\frac{2}{3}}\]
\[a^2=(x+h)^{\frac{4}{3}}\]
\[b=x^{\frac{2}{3}}\]
\[b^2=x^{\frac{4}{3}}\] and so
\[a^2+ab+b^2=(x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}}\]

Answer 37

that is what you have to multiply the top and bottom by. when you multiply in the top you just get
\[a^3-b^3=(x+h)^2-x^2=x^2+2xh+h^2-x^2=2xh+h^2\]

Answer 38

when you multiply in the denominator you get
\[h((x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}})\]

Answer 39

I don't get where multiply the top and bottom

Answer 40

now you are good to go. factor an h out of the numerator, cancel with that factored h in the denominator and you get
\[\frac{2x+h}{(x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}}}\]

Answer 41

ok let me write it out

Answer 42

\[\frac{((x+h)^{\frac{2}{3}}-x^{\frac{2}{3}})\times ((x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}})}{h((x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}})}\]

Answer 43

you get
\[\frac{(x+h)^2-x^2}{h((x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}})}\]

Answer 44

giving
\[\frac{2xh+h^2}{h((x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}})}\]
\[=\frac{h(2x+h)}{h((x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}})}\]
\[=\frac{2x+h}{(x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}}}\]

Answer 45

now replace h by 0 to get your answer

Answer 46

so much work, really confusing

Answer 47

i agree this was a pain. i think i have all the steps written out. the algebra is correct i am pretty sure

Answer 48

@staellite73 cant we just binomialy expand the numerator and cancell out h form numerator and denominator

Answer 49

like Vicky thought can we do that?

Answer 50

sure if you know the binomial expansion for fractional exponents, which is basically a taylor series which requires knowing derivatives.

Answer 51

if you have an elementary method i would be happy. but i think you really need the difference of two cubes and while i agree that the algebra is annoying, it is just algebra

Answer 52

on the numerator did you just cancel out to get (x+h)^2-x^2

Answer 53

@ta etal i am fairly confident in the algebra and while it is long, it is all there. work through the steps and i think you will get it. if your eyes glaze over like mine do when you see such long steps, there really is not much else to do

Answer 54

that is right.

Answer 55

if by "cancel" you mean that
\[((x+h)^{\frac{2}{3}})^3=(x+h)^2\]

Answer 56

and similarly
\[(x^{\frac{2}{3}})^3=x^2\]

Answer 57

and the denominator I thought h isn't suppose have anything plug into it except 0, but you put the same thing from the numerator on the bottom, why?

Answer 58

if you multiply the top by something you also have to multiply the bottom right? like building up fractions when you add them

Answer 59

I still don't see how you get the 2/(x)^(1/3) from all the work?

Answer 60

did you replace h by zero at the last step?

Answer 61

replace h by zero here
\[=\frac{2x+h}{(x+h)^{\frac{4}{3}}+(x+h)^{\frac{2}{3}}\times x^{\frac{2}{3}}+x^{\frac{4}{3}}}\]

Answer 62

yes

Answer 63

what do you get?

Answer 64

wait I got it

Answer 65

whew you scared me

Answer 66

its, still really long hard process to do this problem

Answer 67

hey i agree

Answer 68

yep that's why the power rule is so handy

Answer 69

there is not always a snap way to do something though

Answer 70

Eventhough the power rule help, its still is tedious problem with a lot of step

Answer 71

well there is in this case of course. and remember that it is only algebra you are using. math gets much harder and much longer believe me

Answer 72

Can get help with one more problem please

Answer 73

power rule:

\[\LARGE \frac{d}{dx}\left(x^n\right)=nx^{n-1}\]


but use this when you're more familiar on how/why this works

Answer 74

not if it is as long as this one!

Answer 75

no its not, lol

Answer 76

the limit is 30 but i have no idea where the equation comes from
post and i bet someone will answer quickly

Answer 77

ok thanks