Question

PLEASE HELP ME!

Answer 1

\[G(t) = (2t+3)^2(3t^2-1)^-3\]

Answer 2

explain how

Answer 3

\[G(t) = (2t+3)^2(3t^2-1)^{-3}\]

Answer 4

What to explain in this??

Answer 5

find derivative

Answer 6

https://www.khanacademy.org/math/differential-calculus/taking-derivatives/product_rule/v/applying-the-product-rule-for-derivatives

Answer 7

@Astefank This is the third problem I have seen you post with no work of your own demonstrated. Please do better than that. Show YOUR work!

Answer 8

there is a reason why I'm posting here. If i could do it why would be here

Answer 9

that is the question

Answer 10

and that link is not very helpful

Answer 11

The purpose of OpenStudy is for you to get help with YOUR work. This is not possible if you show NONE. This is the answer to your question. If you truly have NOTHING to offer, you need more help than can be provided in this setting. Show Your Work!

Did you learn the Quotient Rule or not?
Did you learn the Multiplication Rule or not?
Write them out. Think about them? How do they apply to this problem statement?

Show Your Work.

Answer 12

rude

Answer 13

\[\large \begin{align}G(t) &= (2t+3)^2(3t^2-1)^{-3}\\~\\G'(t) &=\color{red}{\left((2t+3)^2\right)'}(3t^2-1)^{-3} + (2t+3)^2\color{red}{\left((3t^2-1)^{-3} \right)'} \end{align}\]

Answer 14

thank you

Answer 15

i had that tho...

Answer 16

that tick on top right corner is another convenient way to represent a derivative

Answer 17

i know

Answer 18

Honest and Direct often are taken to be rude. This is a gross miscaharacterization. Show Your Work. Ganshie8 just did it all for you.

If you had that, why did you not SHOW us? Show your work.

Answer 19

because its on my paper

Answer 20

good, find the derivative and plug them in. Notice that chain rule is doing its job hidden in the background

Answer 21

ok so when i find that will that be my final answer

Answer 22

yes, just simplify if its possible

Answer 23

but i feel you're very far from the final answer, can you work below :

\(\large \color{red}{\left((2t+3)^2\right)'} = ? \)

\(\large \color{red}{\left((3t^2-1)^{-3} \right)'} = ?\)

Answer 24

2(2t+3)
3(3t^2-1)^-5

Answer 25

not quite, you need to use chain rule

Answer 26

\(\large \color{red}{\left((2t+3)^2\right)'} = 2(2t+3) \color{red}{\left(2t+3\right)'}\)

Answer 27

why use chain rule

Answer 28

because the base is not just `t`, its `2t+3`

Answer 29

ok

Answer 30

the derivative of \(\large t^n\) is indeed \(nt^{n-1}\)
however if it is anything other than \(t\), you need to use chain rule

Answer 31

have you learned chain rule yet ?

Answer 32

yes its just 5:12 in the morning and I'm trying to finish this

Answer 33

you may use wolframalpha if you just want the answer..
this site is really for helping you learn and do the problems on your own.. not just for giving answers :)

Answer 34

ok thank you for your help

Answer 35

http://www.wolframalpha.com/input/?i=%28%282t%2B3%29%5E2%283t%5E2-1%29%5E%28-3%29%29%27

Answer 36

tkhunny is saying true and that is hurting the asker.. :P