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August 1, 2014

August 1, 2014

Posted by **michelle** on Monday, October 8, 2012 at 9:00pm.

y= (1+3x)^12 , (0,1)

and find y' and y"

y=cos(x^2)

I got y' = -2xsin(x^2)

i keep trying to get double prime but everything i try isnt working.

- calculus -
**Reiny**, Monday, October 8, 2012 at 10:41pmfirst:

dy/dx = 12(1+3x)^11 (3)

= 36(1+3x)^11

when x = 0

dy/dx = 36(1)^11 = 36

and of course (0,1) is the y-intercept, so equation is

y = 36x + 1

2nd:

y = cos(x^2)

y' = 2x(-sin(x^2) = -2x sin(x^2)

use the product rule for y''

y'' = (-2x)cos(x^2) (2x) + sin(x^2) ( -2)

= -4x^2 cos(x^2) - 2sin(x^2)

- calculus -
**michelle**, Monday, October 8, 2012 at 11:26pmI've tried entering in 36x+1 earlier and it told me it was incorrect. I thought I did it wrong. Thanks for getting back to me.

- calculus -
**michelle**, Monday, October 8, 2012 at 11:29pmwell i entered it in again and now it took that first answer. i think theres a glitch with the program but thank you for showing me how you got the answers. it helps so much!

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