Question

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

Short answer

The derivative \(F'(x)\) of the written function is \(F'(x) = \sqrt{\sin(x)} * \cos(x)\).

Step-by-step solution

Apply the Fundamental Theorem of Calculus

According to the Fundamental theorem of Calculus Part 1, the derivative of an integral of a function from a to x is the function evaluated at x. Meaning that, if \(F(x)=\int_{a}^{x} f(t) dt\), then \(F'(x)= f(x)\). However, this problem involves an integral with a variable upper limit, \(\sin(x)\), rather than just x, so the Chain Rule for derivatives is also required. The Chain rule states that the derivative of the composition of two functions is the derivative of the outer function multiplied by the derivative of the inner function. Hence, \(F'(x) = f(g(x))g'(x)\).

Apply the Function and the Chain Rule

Here, the function is \(f(t) = \sqrt{t}\) and the upper limit function is \(g(x) = \sin(x)\). According to the Chain Rule, the derivative would be \(f(g(x)) * g'(x)\). Substituting the given functions produces: \(F'(x) = \sqrt{\sin(x)} * \cos(x)\). Thus, the derivative of \(F(x)=\int_{0}^{\sin x} \sqrt{t} dt\) is \(F'(x) = \sqrt{\sin(x)} * \cos(x)\).