Find F '(x) for F(x) = integral[x^3 to 1](cos(t^4)dt)

a. cos(x^7)
b. -cos(x^12)
c. -3x^2cos(x^12)
d. cos(1) - cos(x^12)

asked by Jimmy
  1. By the 2nd FTC,
    F' = -cos((x^3)^4)(3x^2) = -3x^2cos(c^12)

    why minus?

    posted by Steve

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