To solve the problem, we need to find the derivative of the function ( A(x) ) with respect to ( x ) and then evaluate it at ( x = 0 ) and ( x = \frac{\pi}{2} ). The function ( A(x) ) is defined as: [ A(x) = \int_1^x \sqrt{\cos^2 t 1} , dt ] Step 1: Differentiate ( A(x) ) using the Fundamental Theorem of Calculus The Fundamental Theorem of Calculus Part 1 states that if ( F(x) = \int_a^x f(t) , dt ), then ( F'(x) = f(x) ). Applying this to our function: [ A'(x) = \sqrt{\cos^2 x 1} ] Step 2: Evaluate ( A'(x) ) at ( x = 0 ) Substitute ( x = 0 ) into ( A'(x) ): [ A'(0) = \sqrt{\cos^2 0 1} ] Since ( \cos 0 = 1 ): [ A'(0) = \sqrt{1^2 1} = \sqrt{2} ] Step 3: Evaluate ( A'(x) ) at ( x = \frac{\pi}{2} ) Substitute ( x = \frac{\pi}{2} ) into ( A'(x) ): [ A'(\frac{\pi}{2}) = \sqrt{\cos^2 \frac{\pi}{2} 1} ] Since ( \cos \frac{\pi}{2} = 0 ): [ A'(\frac{\pi}{2}) = \sqrt{0^2 1} = 1 ] Conclusion The derivative of ( A(x) ) evaluated