Question

Find \(F^{\prime}(x)\). $$ F(x)=\int_{0}^{x^{2}} \sin \theta^{2} d \theta $$

Short answer

\( F^{\prime}(x) = 8x^{4}\cos(x^{4}) \)

Step-by-step solution

Invoke Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function \( F \) is defined by an integral with a variable limit, that is \( F(x)=\int_{a}^{x^{2}} f(t) dt \) then the derivative of \( F \) is just the original function but the upper limit of the integral replaces the variable of integration in the function. Hence, the first step is to apply this theorem to our function \( F(x) \). So, \( F^{\prime}(x)= f(x^2) = \sin(x^{4}) \).

Apply the Chain Rule

Note that the output from the Fundamental Theorem is \( f(x^2) \), this indicates that a chain rule should be applied. The chain rule, which states that the derivative of a composition of functions is the derivative of the outer function multiplied by the derivative of the inner function, leads us to \( F^{\prime}(x)= f^{\prime}(x^2) \cdot 2x \). Here, \( f^{\prime}(x) \) can be obtained as the derivative of \( \sin(x^{4}) \) which is \(4x^{3}\cos(x^{4}) \).

Compute the Derivative

Substituting \( f^{\prime}(x^2) \) to our chain rule we derived, we will have \( F^{\prime}(x)= 4x^{3}\cos(x^{4}) \cdot 2x = 8x^{4}\cos(x^{4}) \). Therefore, the derivative of \( F(x) \) is \( 8x^{4}\cos(x^{4}) \).