If`f(x)=|x|^(|sinx|)`,then`f'((pi)/(4))` equals

Options

A. `((pi)/(4))^(1sqrt2).((sqrt2)/(2)log"(4)/(pi)-(2sqrt2)/(pi))`
B. `((pi)/(4))^(1sqrt2).((sqrt2)/(2)log"(4)/(pi)+(2sqrt2)/(pi))`
C. `((pi)/(4))^(1sqrt2).((sqrt2)/(2)log"(4)/(pi)-(2sqrt2)/(pi))`
D. `((pi)/(4))^(1sqrt2).((sqrt2)/(2)log"(4)/(pi)+(2sqrt2)/(pi))`

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = |x|^{|\sin x|} \) at \( x = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Simplify the function in the neighborhood of \( x = \frac{\pi}{4} \)
Since \( \frac{\pi}{4} \) is positive, we have:
\[ f(x) = x^{\sin x} \]

### Step 2: Take the natural logarithm of both sides
Let \( y = f(x) \). Then,
\[ \log y = \log(x^{\sin x}) = \sin x \cdot \log x \]

### Step 3: Differentiate both sides with respect to \( x \)
Using implicit differentiation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x \cdot \log x) \]
Using the product rule on the right-hand side:
\[ \frac{d}{dx}(\sin x \cdot \log x) = \cos x \cdot \log x + \sin x \cdot \frac{1}{x} \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \log x + \frac{\sin x}{x} \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) \]

### Step 5: Substitute back \( y = x^{\sin x} \)
Thus,
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) \]

### Step 6: Evaluate at \( x = \frac{\pi}{4} \)
Now, we need to evaluate \( f'(\frac{\pi}{4}) \):
1. Calculate \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
2. Calculate \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
3. Calculate \( \log \frac{\pi}{4} \)

Substituting these values into the derivative:
\[ f'\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^{\frac{1}{\sqrt{2}}} \left( \frac{1}{\sqrt{2}} \cdot \log \frac{\pi}{4} + \frac{1/\sqrt{2}}{\frac{\pi}{4}} \right) \]

### Step 7: Simplify the expression
This simplifies to:
\[ f'\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^{\frac{1}{\sqrt{2}}} \left( \frac{1}{\sqrt{2}} \log \frac{\pi}{4} + \frac{4}{\sqrt{2} \pi} \right) \]

### Final Result
Thus, the value of \( f'(\frac{\pi}{4}) \) is:
\[ f'\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^{\frac{1}{\sqrt{2}}} \left( \frac{1}{\sqrt{2}} \log \frac{\pi}{4} + \frac{4}{\sqrt{2} \pi} \right) \]

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