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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