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Calculus1 15 Online
OpenStudy (anonymous):

find first and second derivative of t^2(t-2)

OpenStudy (kainui):

So have you heard of the product rule? You'll need to use that.

OpenStudy (anonymous):

first *derivative of the second +second*derivative of the first

OpenStudy (anonymous):

i need help on starting both derivatives

OpenStudy (anonymous):

if you could possibly tell me how to start them

OpenStudy (anonymous):

do you know the power rule?

OpenStudy (anonymous):

not all too well i need a refresher

OpenStudy (anonymous):

cool, it states that dy/dx of x^n=nx^(n-1). Do you understand the eq.?

OpenStudy (anonymous):

well is it true that the first derivative: A'(x)= x^2(x-2)?

OpenStudy (anonymous):

no you have to use it with the product rule. so first (x^2) times derivative of second(x-2)+second*derivative of the first (x^2)

OpenStudy (anonymous):

so, A'(x)= (x^2)(x-2)+(x^2/2)?

OpenStudy (anonymous):

not quite

OpenStudy (anonymous):

im sorry about this, i just cant get this chapter right :/

OpenStudy (anonymous):

so you have x^2 using x^n=nx^(n-1). u=you have 2x^2-1=2x

OpenStudy (anonymous):

so then what would be the second derivative A''(x)?

OpenStudy (anonymous):

that not even the complete first derivative :/

OpenStudy (anonymous):

oh my bad, how do you finish? add (x-2)

OpenStudy (anonymous):

here I'll show you

OpenStudy (anonymous):

okay thank you

OpenStudy (anonymous):

first is :x^2 serivative of first:2x second is (x-2) derivative of second is: 1

OpenStudy (anonymous):

plugging that in to the product rule:x^2(1)+(x-2)(2x)

OpenStudy (anonymous):

now you have to do it again but twice for each multiple

OpenStudy (anonymous):

well actually only once for (x-2)(2x)

OpenStudy (anonymous):

i hate to ask, but what would be the second derivative? im still having trouble figuring it out

OpenStudy (anonymous):

the second derivative is the derivative of the first derivative

OpenStudy (anonymous):

Here watch this entire series it helped me get through calc one, it's long but it's worth it: http://www.youtube.com/watch?v=ay8838UZ4nM

OpenStudy (anonymous):

so, (2x)(1)(2)= 4x?

OpenStudy (anonymous):

not quite write out what first,second, the derivative of the first, and the derivative of the second are.

OpenStudy (anonymous):

the (2x) and(1)?

OpenStudy (anonymous):

fill this out for the product (x-2)(2x) first is : derivative of first: second is: derivative of second is:

OpenStudy (anonymous):

then plug it into:first *derivative of the second +second*derivative of the first

OpenStudy (anonymous):

ok...

OpenStudy (anonymous):

Here its easier if you simplify x^2(1)+(x-2)(2x) into x^2+2x^2-4x

OpenStudy (anonymous):

so you have 3x^2-4x do the power rule on that

OpenStudy (anonymous):

could you possibly remind me of the power rule?

OpenStudy (anonymous):

x^n=nx^(n-1)

OpenStudy (anonymous):

so 6x-4= 2(3x-3)

OpenStudy (anonymous):

*2(3x-2)

OpenStudy (anonymous):

Yeah!

OpenStudy (anonymous):

THANK YOU!

OpenStudy (anonymous):

np :)

OpenStudy (anonymous):

which of the two graphs would i look at to determine the concavity of the function?

OpenStudy (anonymous):

2nd

OpenStudy (anonymous):

thank you so much

OpenStudy (anonymous):

np

OpenStudy (anonymous):

also, sorry to bother you, but what is the chain rule exactly?

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