Question

Tips when finding the second derivative, I always get lost...

Answer 1

Got any specific questions you want to try?

Answer 2

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

Answer 3

if given y, find dy/dx to find the first derivative, imagine dy/dx as a new function and do the same to find d^y/dx^2

Answer 4

use chain rule, then your going to have to use product/chain rules again for second derivative

just have to write it neatly and keep track of everything

Answer 5

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

Answer 6

So we are going to differentiate that twice? Awesome, let's do this.

First things first, it's extremely helpful to convert the cube root into an exponent. What would we do turn a cube root into an exponent?

Answer 7

there is a 3 in the denominator

Answer 8

Let's do the first one first, and then we will do the second one next.

Answer 9

need to multiply by 1/3

Answer 10

Or do you want to just solve the second problem instead of the first?

Answer 11

Funinabox: that's not how you do it. a 1/3 wouldn't undo a cube root.

Answer 12

when taking the derivative using the power rule, you have to multiply by the exponent and bring the function to a power of exponent - 1

that was all done, except she never multiplied by the exponent (1/3)

Answer 13

\[\frac{ 3(x^2-1)^{-2/3}(2)-2x(-2(x^2-1)^{-2/3})(-2x) }{ 3\sqrt[3]{(x^2-1)^2}^2 }\]

Answer 14

because cuberoot(x^2 - 1) is (x^2 - 1)^1/3

Answer 15

Oh, my bad!

I thought Calle was posting a new problem and that you were suggesting multiplying by 1/3 to undo the cuberoot. Lolz.

She/he never really said whether they were trying to solve it or posting a new problem. :P

Answer 16

i.. think that is correct, but there should be a 9 on the bottom now. instead of 3. as you had to do 1/3 twice.

Answer 17

But yeah, Calle: if you're reading this, we should go back to the beginning and start from scratch to be safe.

Doing it step by step would help us catch errors. Care to do it one step at a time?

Answer 18

sure

Answer 19

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

(1/3)(x^2-1)^(-2/3) * 2x
So that first part is doing the first part of chainrule, the 2x is doing the second part of chain rule (the inside).

Do we all agree that this is good so far? Before I begin simplifying it.

Answer 20

yep that's right

Answer 21

Yay! So that turns into this (sorry, not sure how to make fractions on this site)

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

So far so good, I hope?

Answer 22

(I'll assume it is and begin working on the second derivative :P)

Answer 23

f(x) = 2x
f'(x) = 2

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

g'(x) = Oh dear heaven... This deserves its own comment. Lol.

Answer 24

g(x) = 3*(x^2 - 1) ^(2/3)

Product rule:

f'g + fg'
f = 3
f' = 0
g = (x^2 - 1) ^(2/3)
g' = (2/3)(x^2-1)^(-1/3)*2x

Do we agree? Before I plug these into the product rule?

Answer 25

(I don't like writing f(x), so assume that f means f(x))

Answer 26

in order to use the produce rule you must make the exponent negative

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

should be

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

otherwise you need to use the quotient rule

Answer 27

product rule* lol produce rule, sounds like something in a grocery store

Answer 28

Oh, I was just using the product rule to find out what g'(x) was for when we use the quotient rule. ;)

Answer 29

Basically I was differentiating the denominator with the product rule and will plug that into the quotient rule once we verify that I did g'(x) correctly.

Answer 30

yes but my point is, you can either find the derivative of g(x) by:

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

or

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

Answer 31

that is 1/[3(x^2-1)^(2/3)]

Answer 32

But either way, in the end we would end up using product rule anyway to differentiate it, no?

Answer 33

whether it's for f'x in the product rule, or g'x in the quotient rule.

Answer 34

yes, you can use whatever rule you want, but if your going product rule:

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

and quotient rule:

g(x) = 1[3(x^2-1)^(2/3)]

Answer 35

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

i meant.

reason being is g(x) is a rational function

Answer 36

But let's go ahead and try it out; we should hopefully arrive at the same solution.

f'g + fg'
f = 3
f' = 0
g = (x^2 - 1) ^(2/3)
g' = (2/3)(x^2-1)^(-1/3)*2x

0*(x^2 - 1) ^(2/3) + 3((2/3)(x^2-1)^(-1/3)*2x)

So g'(x) for our quotient rule will be:
3((2/3)(x^2-1)^(-1/3)*2x)

Let's see what happens; we should hopefully get the same answer.

Answer 37

to reiterate:
g= 3(x^2 - 1) ^(2/3)
g' = (2/3)(x^2-1)^(-1/3)*2x

g' is incorrect no matter what, because you either have to start with g being:

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

or

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

because again, g is in the denominator and is a rational function

Answer 38

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

man i keep doing typos, getting confused over what g and what g' are right lol

Answer 39

Hmm... Are you sure about that? I could have sworn that g(x) would only be the stuff on the bottom, so if you have a 3 on the bottom, it'll be treated as a 3 and not a 1/3. Lemme research this to be safe; you may very well be correct, but I don't ever remember doing it your way (except to pull out the 1/3 for an integral or limit).

Answer 40

it is only the stuff on the bottom, but you have to remember - its on the bottom. g(x) is a rational function

its f(x) * 1/g(x)

Answer 41

We're having some miscommunications here:

I am fully aware that we would have to do this if I was using the product rule to solve the second derivative. BUT, I'm using the quotient rule to solve.

So in other words, we started with:

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


Quotient rule is:

f'g - fg'
------
g^2
I'm fairly certain we agree on this (and double checked to make sure this formula was correctly written).

f(x) is 2x. We agree on this.
f'(x) is 2. We agree on this.
g(x) is 3*(x^2 - 1) ^(2/3) We agreed on this earlier.

Here's where we're debating:
g'(x). I am stating that in order to figure out what g'(x) is (for the purpose of QUOTIENT RULE), we have to use the product rule on g(x) so that we can have a value for g'(x). I mean, we are basically arguing as to whether g(x) is whatever is in the denominator, or if g(x) is to be treated as 1/(the stuff in the denominator). I'm saying g(x) = 3*(x^2 - 1) ^(2/3); you're saying g(x) = 1/(g(x)), which as you can see is a paradox.

So...
to solve g'(x), we have to come up with some new functions. Remember, this is ONLY work being done to g(x); we're completely isolating ourselves from the quotient rule that we're working with later. This is just to get g'(x).

f = 3
f' = 0
g = 3*(x^2 - 1) ^(2/3)
g' = (2/3)(x^2 - 1)^(-1/3)*2x

Product rule: (again, only to figure out g'(x)!)

f'g + fg'

0*3*(x^2 - 1) ^(2/3) - 3*(2/3)(x^2 - 1)^(-1/3)*2x
0 - 2x(-1/3)((x^2 - 1)^(-1/3))

g'(x) = 2x
------------
3*(x^2-1)^(1/3)


Sound good? Good.

Answer 42

Back to my previous post:

"So in other words, we started with:

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


Quotient rule is:

f'g - fg'
------
g^2
I'm fairly certain we agree on this (and double checked to make sure this formula was correctly written).

f(x) is 2x. We agree on this.
f'(x) is 2. We agree on this.
g(x) is 3*(x^2 - 1) ^(2/3) We agreed on this earlier."

g'(x) = 2x
------------
3*(x^2-1)^(1/3)

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

Anyhow, that's what I say; let's see if wolfy agrees.

Answer 43

Lolz, it looks like I flipped a sign somewhere; same graph, but upside down.