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 \( \phi(x) = \log_8(\log_3 x) \) at \( x = e \).

### Step-by-Step Solution:

1. **Rewrite the function using change of base formula**:
\[ \phi(x) = \frac{\log(\log_3 x)}{\log(8)} \]
Here, we use the natural logarithm (base \( e \)) for convenience.

**Hint**: Remember that \( \log_b a = \frac{\log_k a}{\log_k b} \) for any base \( k \).

2. **Differentiate \( \phi(x) \)**:
Using the quotient rule, we differentiate \( \phi(x) \):
\[ \phi'(x) = \frac{1}{\log(8)} \cdot \frac{d}{dx}[\log(\log_3 x)] \]
We will need to apply the chain rule to differentiate \( \log(\log_3 x) \).

**Hint**: The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).

3. **Differentiate \( \log(\log_3 x) \)**:
First, we need to find \( \frac{d}{dx}[\log_3 x] \):
\[ \log_3 x = \frac{\log x}{\log 3} \implies \frac{d}{dx}[\log_3 x] = \frac{1}{\log 3} \cdot \frac{1}{x} \]
Now, applying the chain rule:
\[ \frac{d}{dx}[\log(\log_3 x)] = \frac{1}{\log_3 x} \cdot \frac{1}{\log 3} \cdot \frac{1}{x} \]

**Hint**: Keep track of the chain rule and the derivatives of logarithmic functions.

4. **Combine the derivatives**:
Now substituting back into \( \phi'(x) \):
\[ \phi'(x) = \frac{1}{\log(8)} \cdot \left(\frac{1}{\log_3 x} \cdot \frac{1}{\log 3} \cdot \frac{1}{x}\right) \]

5. **Evaluate \( \phi'(e) \)**:
Now we substitute \( x = e \):
\[ \phi'(e) = \frac{1}{\log(8)} \cdot \left(\frac{1}{\log_3 e} \cdot \frac{1}{\log 3} \cdot \frac{1}{e}\right) \]
We know that \( \log_3 e = \frac{1}{\log e} = 1 \) (since \( \log e = 1 \)).

**Hint**: Remember that \( \log_3 e \) can be simplified using the change of base formula.

6. **Final simplification**:
Thus:
\[ \phi'(e) = \frac{1}{\log(8)} \cdot \left(\frac{1}{1} \cdot \frac{1}{\log 3} \cdot \frac{1}{e}\right) = \frac{1}{\log(8) \cdot \log(3) \cdot e} \]

### Final Answer:
\[ \phi'(e) = \frac{1}{\log(8) \cdot \log(3) \cdot e} \]