IDNLearn.com provides a comprehensive solution for all your question and answer needs. Get the information you need from our community of experts who provide accurate and thorough answers to all your questions.
Sagot :
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]
### 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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
