Find the derivative and simplify your answer. \[ \begin{array}{l}A(x)=\int_{2 x}^{1} \sec t d t \\ A^{\prime}(x)=[?][\quad] 2 x\end{array} \]

Given \[ A(x)=\int_{2x}^{1} \sec t \, dt, \] we differentiate with respect to \( x \) using the Leibniz rule for differentiating an integral with variable limits. The Leibniz rule states that if \[ F(x)=\int_{g(x)}^{h(x)} f(t) \, dt, \] then \[ F'(x) = f(h(x))h'(x) - f(g(x))g'(x). \] Here, we identify: - \( f(t)=\sec t \), - Upper limit \( h(x)=1 \) so \( h'(x)=0 \), - Lower limit \( g(x)=2x \) so \( g'(x)=2 \). Substitute into the formula: \[ A'(x)=\sec(1)\cdot 0-\sec(2x)\cdot 2 = -2\sec(2x). \] Thus, the derivative is \[ A'(x)=-2\sec(2x). \]