# calculus

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Find an equation of the tangent line to the curve at the given point.

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 -

first:
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 -

I'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 -

well 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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