The maximum value of `(sin x)^((sin x))` is

Options

A. `7//3`
B. 7
C. `pi//2`
D. 1

Answer

To find the maximum value of the function \( y = (\sin x)^{\sin x} \), we can follow these steps:

### Step 1: Take the logarithm of the function
We start by taking the natural logarithm of both sides:
\[ \log y = \sin x \cdot \log(\sin x) \]

### Step 2: Differentiate both sides
Next, we differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \log(\sin x) + \sin x \cdot \frac{\cos x}{\sin x} \]
This simplifies to:
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \log(\sin x) + \cos x \]
Factoring out \( \cos x \):
\[ \frac{1}{y} \frac{dy}{dx} = \cos x (1 + \log(\sin x)) \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ \cos x (1 + \log(\sin x)) = 0 \]
This gives us two cases:
1. \( \cos x = 0 \)
2. \( 1 + \log(\sin x) = 0 \)

### Step 4: Solve for critical points
**Case 1:** \( \cos x = 0 \) leads to:
\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
**Case 2:** \( 1 + \log(\sin x) = 0 \) leads to:
\[ \log(\sin x) = -1 \implies \sin x = e^{-1} \]

### Step 5: Evaluate the function at critical points
Now we evaluate \( y \) at the critical points.

**At \( x = \frac{\pi}{2} \):**
\[ y = (\sin(\frac{\pi}{2}))^{\sin(\frac{\pi}{2})} = 1^1 = 1 \]

**At \( \sin x = e^{-1} \):**
\[ y = (e^{-1})^{e^{-1}} = e^{-1/e} \]

### Step 6: Compare the values
Now we compare the two values:
1. \( 1 \)
2. \( e^{-1/e} \)

Since \( e^{-1/e} < 1 \) (as \( e^{-1/e} \) is a positive number less than 1), the maximum value occurs at \( x = \frac{\pi}{2} \).

### Conclusion
The maximum value of \( (\sin x)^{\sin x} \) is:
\[ \boxed{1} \]