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

1 answer:

4 0

The upper limit is constant, so the derivative of the integral evaluated there is 0. The derivative of the integral evaluated at the lower limit is the product of the integrand evaluated at the lower limit and the derivative of the lower limit.
F'(x) = 0 -cos((x^3)^4)·3x^2
Simplifying, ...
F'(x) = -3x^2·cos(x^12)

The appropriate choice is ...
B. -3x^2cos(x^12)

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