Question

anyone have some good hints when finding the second derivative, I always get lost :(

Answer 1

With what question?

Answer 2

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

Answer 3

Well I would make that (x^2-1)^(1/3)

Answer 4

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

Answer 5

Then take the derivative combining the power rule and the chain rule.

Answer 6

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

Answer 7

Yeah that's right.

Answer 8

\[\frac{ 3(x^2-1)^{-2/3}(2)-2x(-2(x^2-1)^{-4/3}(2x) }{ 3\sqrt[3]{(x^2-1)^2}^2 }\]

Answer 9

*faints*

Answer 10

Wait.

Answer 11

Do you need the second derivative, or do you need just tips?

Answer 12

well, i can look up the second derivative, im just wondering if there is an effective way to solve this, i always get lost

Answer 13

Not exactly, It just comes with practice. Eventually when you get to calc iii, it starts becoming second nature. That's because for Diff. Eq., you must understand them well enough to work backwards. Sorry for no helpful tips.

Answer 14

Consistent notation helps a little bit.

Like I notice that in the bottom you're using a root symbol, and in the top you're using fractional exponents. I wouldn't mix those like that unless you're really comfortable with it.
Personally I tend to use the check mark if its a square root, but write a fraction if it's any other power/root.

I've seen you do a lot of these problems, you seem pretty great with your algebra and middle steps. I would just recommend getting really comfortable with the product and quotient rule.

Answer 15

\[\frac{ 6 }{ \sqrt[3]{x^2-1} }+\frac{ 8x^2 }{ \sqrt[3]{(x^2-1)^5} }\]

Answer 16

thats all over

Answer 17

ew you gave them the check marks XD lol

Answer 18

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

Answer 19

thanks zep, i will work on it more :)

Answer 20

so, can i cross cancel the top portion?

Answer 21

Might wanna back up a lil bit and check your work :c hmm

Answer 22

checking