Question

What is the derivative of x^(2/3)(x+2)

Answer 1

\[X ^{2/3}(x+2)\]

Answer 2

@amistre64

Answer 3

@phi @robtobey

Answer 4

did you try, using the product rule ?

Answer 5

No! I forgot!! LOL

Answer 6

Thanks!!!!!

Answer 7

or, distribute the x^(2/3) and find the derivative of
\[ x^\frac{5}{3} + 2x^\frac{2}{3} \]

Answer 8

Ok, how do you do that? I made a mistake somewhere when I was doing that

Answer 9

I brought the exponents down and subtracted the powers by 1

Answer 10

write x^1 as x^(3/3)
x^(2/3) * x^(3/3) = x^(2/3 + 3/3) i.e. add exponents

Answer 11

Where'd you get x^1??

Answer 12

adding exponents is easy to remember if you remember
x^1 * x^2 (for example) is x*x*x= x^3

Answer 13

Oh, yeah

Answer 14

x^(2/3)(x+2)

Answer 15

Ok

Answer 16

distribute the x^2/3

x^2/3 * x + 2 x^(2/3)

Answer 17

So x^5/3+4/3x^-1/3

Answer 18

Oh whoops

Answer 19

Its 5/3x^2/3+4/3x^-1/3

Answer 20

yes, that is one way to write it.

Answer 21

Ok, now how do I simplify that

Answer 22

I need to find the critical points

Answer 23

set it equal to 0

Answer 24

Well to do that, I would like to simplify it further. Thats where I was making my mistakes

Answer 25

I got (5x^2/3+4)/(3cuberoot(x)), but there is not supposed to be an exponent in the numerator

Answer 26

\[ \frac{5}{3} x^\frac{2}{3} + \frac{4}{3} x^{-\frac{1}{3} } = 0 \]
multiply both sides by 3
\[ 5 x^\frac{2}{3}+ 4 x^{-\frac{1}{3} } = 0 \]
multiply both sides by x^(1/3)

Answer 27

What I meant was to convert the -1/3 to 1/cube root of x

Answer 28

Its easier for me that way because I can set the numerator equal to 0 to find the places where the derivative is 0, and so I can easily set the denominator equal to 0 to find the places where the derivative is undefined

Answer 29

I would try the way I am suggesting

Answer 30

Can we try it my way? I don't want to get confused and I like to be able to tell where the derivatives 0s are and where its undefined

Answer 31

when working with fractional exponents, it is easiest to keep them as exponents. In other words, do not write it as cube root.
starting with the derivative set equal to 0:
\[ \frac{5}{3} x^\frac{2}{3} + \frac{4}{3} x^{-\frac{1}{3} } = 0 \]
what did you do ? Please go in small steps so I can follow.

Answer 32

btw, the only way to get a "divide by 0" is when x=0.

Answer 33

I did \[((5x ^{2/3})/3)+4/(3\sqrt[3]{x})\]

Answer 34

set equal to 0

Answer 35

Actually, I'm not sure how I got that

Answer 36

but you really should write it as x^(1/3) not cube root.

Answer 37

Why? Shouldn't I combine like terms first

Answer 38

now what ?

Answer 39

I want to just have a numerator over a denominator, not two different fractions

Answer 40

you need a common denominator

Answer 41

Ok

Answer 42

5x/3cube root of x + 4/3 cube root of x

Answer 43

Ooooooh

Answer 44

I don't think I found a common denominator. I think i just saw the two fractions (4/3 and 5/3) and said that 3 will be the denominator

Answer 45

as long as x is not 0, you can ignore the bottom
and solve for the top=0

Answer 46

Yeah I get it now! Thanks a lot!!! These careless mistakes kill!

Answer 47

fyi, if you did this
\[ 5 x^\frac{2}{3}+ 4 x^{-\frac{1}{3} } = 0 \\
5 x^\frac{2}{3}x^\frac{1}{3}+ 4 x^{-\frac{1}{3} }x^\frac{1}{3} = 0
\\5x + 4 =0 \\ x = - \frac{4}{5}
\]