The derivative of $\int_{\sin x}^{\cos x} t^{2} d t$ is $\sin ^{2} x-\cos ^{2} x$ $-\cos x \sin ^{2} x-\sin x \cos ^{2} x$ $\cos x \sin ^{2} x+\sin x \cos ^{2} x$ 1 $\cos ^{2} x-\sin ^{2} x$

Step 1 ：The problem is asking for the derivative of a definite integral with variable limits. This is a direct application of the Leibniz rule (also known as the Fundamental Theorem of Calculus). The Leibniz rule states that the derivative of an integral of a function with respect to its upper limit minus the derivative of the integral with respect to its lower limit is equal to the function evaluated at the upper limit minus the function evaluated at the lower limit. In this case, the function is \(t^2\), the upper limit is \(\cos x\) and the lower limit is \(\sin x\). So, we need to calculate the derivative of \(t^2\) evaluated at \(\cos x\) and \(\sin x\) and subtract the two.

Step 2 ：The derivative of \(\sin^2 x - \cos^2 x\) is \(4\sin x \cos x\). However, this does not match any of the options given in the question. It seems there might be a mistake in my calculations. Let's try to apply the Leibniz rule directly to the integral.

Step 3 ：The derivative of the integral is \(-\sin^2 x \cos x - \sin x \cos^2 x\). This matches the second option in the question. Therefore, the derivative of \(\int_{\sin x}^{\cos x} t^{2} d t\) is \(-\sin x \sin^2 x - \sin x \cos^2 x\).

Step 4 ：\(\boxed{-\sin x \sin^2 x - \sin x \cos^2 x}\) is the final answer.