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)

By the 2nd FTC,

F' = -cos((x^3)^4)(3x^2) = -3x^2cos(c^12)

why minus?

To find the derivative of F(x), we can use the Fundamental Theorem of Calculus. According to this theorem, if F(x) is defined as the accumulation (integral) of a function f(t) over an interval [a, x], then F'(x) is equal to f(x).

In this case, we have F(x) = ∫[x^3 to 1] (cos(t^4) dt. To find F'(x), we need to find the derivative of the integral with respect to x.

To apply the Fundamental Theorem of Calculus, we rewrite the integral using the upper limit x. The integral becomes:

F(x) = ∫[1 to x^3] (cos(t^4) dt

Now, to differentiate F(x), we can use the chain rule since the upper limit is dependent on x. The chain rule states that if we have an integral with limits that are functions of x, we differentiate the integrand with respect to the upper limit (x^3 in this case) multiplied by the derivative of the upper limit.

Differentiating f(x) = cos(t^4) with respect to t gives f'(t) = -4t^3sin(t^4). We can now substitute this into our differentiation:

F'(x) = f(x^3) * (d/dx) (x^3)

Applying the chain rule to (d/dx)(x^3) = 3x^2, we have:

F'(x) = f(x^3) * 3x^2

F'(x) = (-4(x^3)^3)sin((x^3)^4) * 3x^2

Simplifying further,

F'(x) = -12x^8sin(x^12) * 3x^2

F'(x) = -36x^10sin(x^12)

Hence, the derivative of F(x) is given by -36x^10sin(x^12).

Comparing the result to the answer choices, we see that none of the options matches the derivative we found. Therefore, none of the given options is correct.