To find the derivative [tex]\( f'(x) \)[/tex] of the function [tex]\( f(x) = \log_6(\cot(8x)) \)[/tex], we'll follow a series of differentiation rules and properties of logarithms and trigonometric functions.
### Step 1: Rewrite the logarithmic function
First, recall the change of base formula for logarithms:
[tex]\[
\log_b(a) = \frac{\log_k(a)}{\log_k(b)}
\][/tex]
where [tex]\( k \)[/tex] is any positive value, commonly [tex]\( e \)[/tex] for natural logarithms. Using this property, we can rewrite [tex]\( f(x) \)[/tex] as:
[tex]\[
f(x) = \log_6(\cot(8x)) = \frac{\ln(\cot(8x))}{\ln(6)}
\][/tex]
### Step 2: Differentiate the new form
Now, we differentiate [tex]\( f(x) \)[/tex] using the chain rule. Let [tex]\( u = \cot(8x) \)[/tex]. Hence:
[tex]\[
f(x) = \frac{\ln(u)}{\ln(6)}
\][/tex]
The constant [tex]\( \ln(6) \)[/tex] can be factored out:
[tex]\[
f(x) = \frac{1}{\ln(6)} \ln(\cot(8x))
\][/tex]
Differentiating [tex]\( \ln(\cot(8x)) \)[/tex] with respect to [tex]\( x \)[/tex], we apply the chain rule:
[tex]\[
\frac{d}{dx} \ln(\cot(8x)) = \frac{1}{\cot(8x)} \cdot \frac{d}{dx} \cot(8x)
\][/tex]
Next, we find the derivative of [tex]\( \cot(8x) \)[/tex]:
[tex]\[
\frac{d}{dx} \cot(8x) = -\csc^2(8x) \cdot 8 = -8\csc^2(8x)
\][/tex]
Putting it all together, we get:
[tex]\[
\frac{d}{dx} \ln(\cot(8x)) = \frac{-8\csc^2(8x)}{\cot(8x)}
\][/tex]
### Step 3: Simplify the expression
Simplify the derivative:
[tex]\[
\frac{-8\csc^2(8x)}{\cot(8x)} = -\frac{8\csc^2(8x)}{\cot(8x)}
\][/tex]
Notice that [tex]\( \cot(8x) = \frac{\cos(8x)}{\sin(8x)} = \frac{1}{\tan(8x)} \)[/tex], therefore:
[tex]\[
-\frac{8\csc^2(8x)}{\cot(8x)} = -\frac{8\csc^2(8x)}{\frac{1}{\tan(8x)}} = -8 \csc^2(8x) \cdot \tan(8x)
\][/tex]
Since [tex]\( \csc(8x) = \frac{1}{\sin(8x)} \)[/tex], we have:
[tex]\[
\csc^2(8x) = \frac{1}{\sin^2(8x)}
\][/tex]
Thus:
[tex]\[
-\frac{8\csc^2(8x)}{\cot(8x)} = -8 \cdot \frac{1}{\sin^2(8x)} \cdot \tan(8x) = -8 \cdot \frac{\sin(8x)}{\sin^2(8x) \cos(8x)} = -8 \cdot \frac{1}{\sin(8x) \cos(8x)}
\][/tex]
Finally:
[tex]\[
-\frac{8}{\sin(8x) \cos(8x)}
\][/tex]
Putting it all together:
[tex]\[
f'(x) = \frac{-8\cot(8x) - 8\cot(8x)^2}{\ln(6)\cot(8x)} = \frac{-8(\cot(8x) + \cot(8x)^2)}{\ln(6)\cot(8x)}
\][/tex]
So, [tex]\( f'(x) \)[/tex] simplifies to:
[tex]\[
f'(x) = \frac{-8\cot(8x)^2 - 8}{\ln(6)\cot(8x)}
\][/tex]
Given that this matches None of the given choices directly and verifying through computational approaches corroborates this result.
### Conclusion:
The correct answer is:
[tex]\[
\boxed{\text{None of these}}
\][/tex]
