Question

What is d/dx of cos^2 (x^3) ?

Answer 1

\[d/dx [\cos^2(x^3)]\]

Answer 2

Use chain rule and try :)

Answer 3

Ohh wooaah [x3]

You know you love me, I know you care
Just shout whenever, and I'll be there
You are my love, you are my heart
And we will never ever ever be apart

Are we an item? Girl, quit playing
We're just friends, what are you saying?
Say there's another and look right in my eyes
My first love broke my heart for the first time
And I was like...

Baby, baby, baby oooh
Like baby, baby, baby nooo
Like baby, baby, baby oooh
I thought you'd always be mine (mine)

Baby, baby, baby oooh
Like baby, baby, baby nooo
Like baby, baby, baby oooh
I thought you'd always be mine (mine)

Oh, for you I would have done whatever
And I just can't believe we ain't together
And I wanna play it cool, but I'm losin' you
I'll buy you anything, I'll buy you any ring
And I'm in pieces, baby fix me
And just shake me 'til you wake me from this bad dream
I'm going down, down, down, down
And I just can't believe my first love won't be around

And I'm like
Baby, baby, baby oooh
Like baby, baby, baby nooo
Like baby, baby, baby oooh
I thought you'd always be mine (mine)

Baby, baby, baby oooh
Like baby, baby, baby nooo
Like baby, baby, baby oooh
I thought you'd always be mine (mine)

[Ludacris:]
Luda! When I was 13, I had my first love,
There was nobody that compared to my baby
and nobody came between us or could ever come above
She had me going crazy, oh, I was star-struck,
she woke me up daily, don’t need no Starbucks.
She made my heart pound, it skipped a beat when I see her in the street and
at school on the playground but I really wanna see her on the weekend.
She knows she got me dazing cause she was so amazing
and now my heart is breaking but I just keep on saying...

Baby, baby, baby oooh
Like baby, baby, baby nooo
Like baby, baby, baby oooh
I thought you'd always be mine (mine)

Baby, baby, baby oooh
Like baby, baby, baby nooo
Like baby, baby, baby oooh
I thought you'd always be mine (mine)

I'm gone (Yeah Yeah Yeah, Yeah Yeah Yeah)
Now I'm all gone (Yeah Yeah Yeah, Yeah Yeah Yeah)
Now I'm all gone (Yeah Yeah Yeah, Yeah Yeah Yeah)
Now I'm all gone (gone, gone, gone...)
I'm gone

Answer 4

Please explain how 2 differentiate using chain rule....i can never do it correcily

Answer 5

\[d/dx= 2cosx^{3}(-sinx^{3})3x^{2}\]

Answer 6

can u explain the chain rule?

Answer 7

i mean thats not correct

Answer 8

-3 x^2 sin(2 x^3)

Answer 9

sorry the 9 is supposed 2 be a parenthesis. it should be -6x^2 (sinx^3) *cos(x^3)

Answer 10

\[d/dx(\cos^{2}(x^{3}))=2\cos(x^{3})(d/dx(\cos(x^{3})\]

Answer 11

\[2\cos(x^{3})(-\sin(x^{3})(d/dx(x^{3}))\]

Answer 12

\[-2\cos(x^{3})\sin(x^{3})3x^{2}\]

Answer 13

\[-6x^{2}\sin(x^{3})\cos(x^{3})\]

Answer 14

why did the2cos(s^3) turn negative to -2cos(x^3)

Answer 15

because the (-) on the sine can be factored out

Answer 16

please expalin the gist of the chain rule. taht is what u used right?

Answer 17

Chain rule should be applied like.

Eg

Differentiate y =cos(sin(tan(cot(x))))

Now we should assume sin(tan(cot(x)) as t

y = cos t
dy/dt = -sin t * (dt/dx)

Now continue like that

Answer 18

so for my problem it would be like y = cos x and dy/dx = -sin (x^3)....idk chain rule is confusing...can u esplain it using the problem i posted

Answer 19

@help_with_math , see this video. It should help you :)
http://www.khanacademy.org/math/calculus/v/the-chain-rule

Answer 20

ok thanks