Question

Find F’(x) for F(x) = the integral from x^2 to 3 of sin(t^2), dt

Answer 1

Explanation:

To find the derivative of the function F(x) with respect to x, we'll use the Fundamental Theorem of Calculus and the Chain Rule. Let's differentiate the integral with respect to its upper limit and then multiply by the derivative of the upper limit (x^2) with respect to x.

Given:

\[ F(x) = \int_{x^2}^{3} \sin(t^2) \, dt \]

Using the Fundamental Theorem of Calculus:

\[ F'(x) = \sin(3^2) \cdot 3' - \sin(x^2) \cdot (x^2)' \]

Simplify:

\[ F'(x) = 9\sin(9) - 2x\sin(x^2) \]

So, the derivative of \( F(x) \) with respect to \( x \) is \( F'(x) = 9\sin(9) - 2x\sin(x^2) \).