The slope of the tangent to the curve ` y = x + ` sin x cos x at x ` = (pi)/( 2)` is

Options

A. 0
B. `(1)/(2)`
C. 1
D. `-(1)/(2)`

Answer

To find the slope of the tangent to the curve \( y = x + \sin x \cos x \) at \( x = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = x + \sin x \cos x \]
To find the slope of the tangent, we need to compute the derivative \( \frac{dy}{dx} \).

### Step 2: Use the product-to-sum identity
Notice that \( \sin x \cos x \) can be rewritten using the double angle identity:
\[ \sin x \cos x = \frac{1}{2} \sin(2x) \]
Thus, we can rewrite \( y \) as:
\[ y = x + \frac{1}{2} \sin(2x) \]

### Step 3: Differentiate \( y \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{1}{2} \sin(2x)\right) \]
The derivative of \( x \) is \( 1 \), and using the chain rule, the derivative of \( \frac{1}{2} \sin(2x) \) is:
\[ \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x) \]
So, we have:
\[ \frac{dy}{dx} = 1 + \cos(2x) \]

### Step 4: Evaluate the derivative at \( x = \frac{\pi}{2} \)
Next, we need to find the slope at \( x = \frac{\pi}{2} \):
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = 1 + \cos\left(2 \cdot \frac{\pi}{2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = 1 + \cos(\pi) \]
Since \( \cos(\pi) = -1 \), we have:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = 1 - 1 = 0 \]

### Conclusion
Thus, the slope of the tangent to the curve at \( x = \frac{\pi}{2} \) is:
\[ \boxed{0} \]

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