To find the derivative \( \frac{dy}{dx} \) of the function \( y = x \sin x \) and evaluate it at \( x = \frac{\pi}{2} \), we will follow these steps: ### Step 1: Differentiate the function We need to use the product rule for differentiation because \( y \) is the product of two functions: \( x \) and \( \sin x \). The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by: \[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \] In our case: - Let \( u = x \) and \( v = \sin x \). - Then, \( \frac{du}{dx} = 1 \) (the derivative of \( x \)) and \( \frac{dv}{dx} = \cos x \) (the derivative of \( \sin x \)). Now, applying the product rule: \[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x \cos x + \sin x \cdot 1 \] Thus, we have: \[ \frac{dy}{dx} = x \cos x + \sin x \] ### Step 2: Evaluate the derivative at \( x = \frac{\pi}{2} \) Now we need to substitute \( x = \frac{\pi}{2} \) into the derivative we found: \[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = \left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) \] We know that: - \( \cos\left(\frac{\pi}{2}\right) = 0 \) - \( \sin\left(\frac{\pi}{2}\right) = 1 \) Substituting these values in: \[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = \left(\frac{\pi}{2}\right) \cdot 0 + 1 = 0 + 1 = 1 \] ### Conclusion: Final Answer Thus, the value of \( \frac{dy}{dx} \) when \( x = \frac{\pi}{2} \) is: \[ \frac{dy}{dx} = 1 \] The correct option is **C. 1**. ### Explanation of Other Options - **A. -π/2**: This option is incorrect because the derivative we calculated does not yield a negative value, and specifically, it does not equal \(-\frac{\pi}{2}\). - **B. -1**: This option is also incorrect. The derivative evaluated at \( x = \frac{\pi}{2} \) is positive and equals 1, not -1. - **D. π/2**: This option is incorrect as well. While \(\frac{\pi}{2}\) is the value of \( x \), it does not represent the value of the derivative at that point. ### Revision Summary - Use the product rule for differentiation when dealing with products of functions. - Evaluate the derivative at the specified point by substituting the value into the derivative expression. - Remember the values of trigonometric functions at key angles (e.g., \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \)). - Double-check calculations to ensure accuracy in evaluating derivatives and substituting values.