Question:
If $f(x)=logₓ(log~x)$, then $f^\prime(x)$ at $x=e$ is
If $f(x)=logₓ(log~x)$, then $f^\prime(x)$ at $x=e$ is
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If $f(x)=logx(log~x)$, then $f(x)$ at $x=e$ is
Updated On: Apr 15, 2026
- $1/e$
- $e$
- $-1/e$
- 0
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Use the change of base formula: $log_x(log~x) = \frac{log(log~x)}{log~x}$.
Step 2: Analysis
Differentiate using the quotient rule: $f'(x) = \frac{(log~x) \cdot \frac{1}{log~x} \cdot \frac{1}{x} - log(log~x) \cdot \frac{1}{x}}{(log~x)^2}$.
Step 3: Evaluation
Simplify: $f'(x) = \frac{1 - log(log~x)}{x(log~x)^2}$. Now, substitute $x=e$.
Step 4: Conclusion
$f'(e) = \frac{1 - log(log~e)}{e(log~e)^2} = \frac{1 - log(1)}{e(1)^2} = \frac{1 - 0}{e} = \frac{1}{e}$.
Final Answer: (a)
Use the change of base formula: $log_x(log~x) = \frac{log(log~x)}{log~x}$.
Step 2: Analysis
Differentiate using the quotient rule: $f'(x) = \frac{(log~x) \cdot \frac{1}{log~x} \cdot \frac{1}{x} - log(log~x) \cdot \frac{1}{x}}{(log~x)^2}$.
Step 3: Evaluation
Simplify: $f'(x) = \frac{1 - log(log~x)}{x(log~x)^2}$. Now, substitute $x=e$.
Step 4: Conclusion
$f'(e) = \frac{1 - log(log~e)}{e(log~e)^2} = \frac{1 - log(1)}{e(1)^2} = \frac{1 - 0}{e} = \frac{1}{e}$.
Final Answer: (a)
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