If `f(x) = |x|^ |tanx|` then `f'( -pi/6)` is equal to

Explanation

To find \( f'(-\frac{\pi}{6}) \) for the function \( f(x) = |x|^{|\tan x|} \), we will follow these steps:

### Step 1: Analyze the function at \( x = -\frac{\pi}{6} \)

Since \( x = -\frac{\pi}{6} \) is negative, we can simplify \( f(x) \) as follows:
\[ f(x) = (-x)^{|\tan x|} = (-x)^{-\tan(-x)} = (-x)^{\tan x} \]
Thus, we have:
\[ f(-\frac{\pi}{6}) = \left(-\left(-\frac{\pi}{6}\right)\right)^{|\tan(-\frac{\pi}{6})|} = \left(\frac{\pi}{6}\right)^{|\tan(-\frac{\pi}{6})|} \]
Since \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \), we have:
\[ |\tan(-\frac{\pi}{6})| = \frac{1}{\sqrt{3}} \]
So,
\[ f(-\frac{\pi}{6}) = \left(\frac{\pi}{6}\right)^{\frac{1}{\sqrt{3}}} \]

### Step 2: Differentiate \( f(x) \)

To differentiate \( f(x) \), we use the logarithmic differentiation:
\[ \ln f(x) = |\tan x| \ln |x| \]
Differentiating both sides with respect to \( x \):
\[ \frac{f'(x)}{f(x)} = \frac{d}{dx} (|\tan x|) \ln |x| + |\tan x| \frac{d}{dx} (\ln |x|) \]
Using the chain rule, we find:
\[ \frac{d}{dx} (|\tan x|) = \text{sgn}(\tan x) \sec^2 x \]
Thus:
\[ \frac{f'(x)}{f(x)} = \text{sgn}(\tan x) \sec^2 x \ln |x| + |\tan x| \cdot \frac{1}{x} \]
Now, multiplying both sides by \( f(x) \):
\[ f'(x) = f(x) \left( \text{sgn}(\tan x) \sec^2 x \ln |x| + |\tan x| \cdot \frac{1}{x} \right) \]

### Step 3: Evaluate \( f'(-\frac{\pi}{6}) \)

At \( x = -\frac{\pi}{6} \):
- \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \) so \( \text{sgn}(\tan(-\frac{\pi}{6})) = -1 \)
- \( f(-\frac{\pi}{6}) = \left(\frac{\pi}{6}\right)^{\frac{1}{\sqrt{3}}} \)

Now substituting into the derivative:
\[ f'(-\frac{\pi}{6}) = \left(\frac{\pi}{6}\right)^{\frac{1}{\sqrt{3}}} \left( -\sec^2(-\frac{\pi}{6}) \ln \left(\frac{\pi}{6}\right) + \left(-\frac{1}{\sqrt{3}}\right) \cdot \frac{1}{-\frac{\pi}{6}} \right) \]
Calculating \( \sec^2(-\frac{\pi}{6}) \):
\[ \sec^2(-\frac{\pi}{6}) = \frac{1}{\cos^2(-\frac{\pi}{6})} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{4}{3} \]
Thus:
\[ f'(-\frac{\pi}{6}) = \left(\frac{\pi}{6}\right)^{\frac{1}{\sqrt{3}}} \left( -\frac{4}{3} \ln \left(\frac{\pi}{6}\right) + \frac{6}{\pi \sqrt{3}} \right) \]

### Final Answer
\[ f'(-\frac{\pi}{6}) = \left(\frac{\pi}{6}\right)^{\frac{1}{\sqrt{3}}} \left( -\frac{4}{3} \ln \left(\frac{\pi}{6}\right) + \frac{6}{\pi \sqrt{3}} \right) \]