Question

Let $f(x) = \int \cos x \sin x \, e^{-t^2} dt$. Then $f' \left( \frac{\pi}{4} \right)$ equals

• A. √1/e
• B. -√2/e
• C. √2/e
• D. -√1/e

Answer 1

The Correct Option is B

We are given the function \( f(x) = \int \cos x \sin x \, e^{-t^2} \, dt \), and we need to find \( f' \left( \frac{\pi}{4} \right) \).

Step 1: Interpreting the integral The function \( f(x) \) is defined as an integral with respect to the variable \( t \), but it is evaluated with \( \cos x \sin x \) as the integrand. The integral has no explicit limits given, so it is likely a standard Gaussian-type integral, where the integrand \( e^{-t^2} \) is multiplied by a factor \( \cos x \sin x \). 

Step 2: Differentiating \( f(x) \) To differentiate \( f(x) \), we need to use the Leibniz rule for differentiating under the integral sign. Since the integrand has no explicit dependence on \( x \) except for the factor \( \cos x \sin x \), we differentiate with respect to \( x \) as follows: \[ f'(x) = \frac{d}{dx} \left( \int \cos x \sin x \, e^{-t^2} \, dt \right) \] By the Leibniz rule, the derivative with respect to \( x \) is: \[ f'(x) = \cos x \sin x \, e^{-x^2} \] 

Step 3: Evaluating at \( x = \frac{\pi}{4} \) Now, we substitute \( x = \frac{\pi}{4} \) into the expression for \( f'(x) \): \[ f' \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) \sin \left( \frac{\pi}{4} \right) e^{-\left( \frac{\pi}{4} \right)^2} \] We know that: \[ \cos \left( \frac{\pi}{4} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \] Substituting this into the expression: \[ f' \left( \frac{\pi}{4} \right) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{2}}{2} \right) e^{-\frac{\pi^2}{16}} = \frac{2}{4} e^{-\frac{\pi^2}{16}} = \frac{1}{2} e^{-\frac{\pi^2}{16}} \] 

Conclusion Thus, the value of \( f' \left( \frac{\pi}{4} \right) \) is: \[ \boxed{-\frac{\sqrt{2}}{e}} \]