If `phi(x)=log_(8)log_(3)x`, then `phi'(e )` is equal to

Options

A. `elog8`
B. `-elog8`
C. `(1)/(elog8)`
D. None of these

Answer

To solve the problem, we need to find the derivative of the function \( \phi(x) = \log_8(\log_3 x) \) and then evaluate it at \( x = e \).

### Step-by-Step Solution:

1. **Rewrite the logarithm**:
Using the change of base formula, we can rewrite \( \log_8(\log_3 x) \):
\[ \phi(x) = \frac{\log(\log_3 x)}{\log(8)} \]
where \( \log \) denotes the natural logarithm (base \( e \)).

2. **Differentiate \( \phi(x) \)**:
Now, we will differentiate \( \phi(x) \):
\[ \phi'(x) = \frac{1}{\log(8)} \cdot \frac{d}{dx}(\log(\log_3 x)) \]
To differentiate \( \log(\log_3 x) \), we will use the chain rule:
\[ \frac{d}{dx}(\log(\log_3 x)) = \frac{1}{\log_3 x} \cdot \frac{d}{dx}(\log_3 x) \]
We also need to differentiate \( \log_3 x \):
\[ \log_3 x = \frac{\log x}{\log 3} \quad \Rightarrow \quad \frac{d}{dx}(\log_3 x) = \frac{1}{\log 3} \cdot \frac{1}{x} \]
Combining these results, we have:
\[ \frac{d}{dx}(\log(\log_3 x)) = \frac{1}{\log_3 x} \cdot \frac{1}{\log 3} \cdot \frac{1}{x} \]

3. **Substituting back into the derivative**:
Now substituting back into our expression for \( \phi'(x) \):
\[ \phi'(x) = \frac{1}{\log(8)} \cdot \frac{1}{\log_3 x} \cdot \frac{1}{\log 3} \cdot \frac{1}{x} \]

4. **Evaluate at \( x = e \)**:
Now we need to evaluate \( \phi'(e) \):
\[ \phi'(e) = \frac{1}{\log(8)} \cdot \frac{1}{\log_3 e} \cdot \frac{1}{\log 3} \cdot \frac{1}{e} \]
We know that \( \log_3 e = \frac{\log e}{\log 3} = \frac{1}{\log 3} \), so:
\[ \phi'(e) = \frac{1}{\log(8)} \cdot \frac{1}{\frac{1}{\log 3}} \cdot \frac{1}{\log 3} \cdot \frac{1}{e} \]
Simplifying this gives:
\[ \phi'(e) = \frac{1}{\log(8)} \cdot \frac{1}{\log 3} \cdot \frac{1}{e} \]

5. **Final result**:
Thus, the final result is:
\[ \phi'(e) = \frac{1}{e \cdot \log(8) \cdot \log(3)} \]