Let $f(x)=\int_{\sqrt[3]{x}}^{x^{5}} \sin (t) d t$ Use the Fundamental Theorem of Calculus to differentiate \[ D_{x}[f(x)]= \]

Step 1 ：Let \(f(x)=\int_{\sqrt[3]{x}}^{x^{5}} \sin (t) dt\)

Step 2 ：Use the Fundamental Theorem of Calculus to differentiate \(D_{x}[f(x)]\)

Step 3 ：The Fundamental Theorem of Calculus states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral of f from a to b is equal to F(b) - F(a).

Step 4 ：In this case, we have a function f(x) defined as an integral from \(\sqrt[3]{x}\) to \(x^{5}\) of \(\sin(t) dt\). To differentiate this, we need to apply the Fundamental Theorem of Calculus, which in this case will involve the chain rule as well, since the limits of integration are not constants but functions of x.

Step 5 ：The derivative of the integral from a(x) to b(x) of f(t) dt is given by: \(f(b(x)) * b'(x) - f(a(x)) * a'(x)\)

Step 6 ：In this case, a(x) = \(\sqrt[3]{x}\), b(x) = \(x^{5}\), and f(t) = \(\sin(t)\). So we need to calculate the derivatives a'(x) and b'(x), and then substitute these values and the values of a(x) and b(x) into the formula above.

Step 7 ：Calculate the derivatives a'(x) and b'(x): a'(x) = \(\frac{1}{3x^{2/3}}\), b'(x) = \(5x^{4}\)

Step 8 ：Substitute these values into the formula: \(D_{x}[f(x)]= -\frac{1}{3}\frac{\sin(\sqrt[3]{x})}{x^{2/3}} + 5x^{4}\sin(x^{5})\)

Step 9 ：\(\boxed{D_{x}[f(x)]= -\frac{1}{3}\frac{\sin(\sqrt[3]{x})}{x^{2/3}} + 5x^{4}\sin(x^{5})}\) is the final answer.