Question:
Derivative of $x^{(x^x)}$ is
Derivative of $x^{(x^x)}$ is
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$d/dx(x^x) = x^x(1+\ln x)$ is a standard derivative useful for nested functions.
Updated On: May 7, 2026
- $x^{(x^x)} (x^x + 1 + \log x)$
- $x^{(x^x)} (x^x + \log x)$
- $x^{(x^x)} (x^x + x^{x-1} \log x (1 + \log x))$
- $x^{(x^x)} (x^{x-1} + x^x \log x (1 + \log x))$
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The Correct Option is D
Solution and Explanation
Step 1: Logarithmic Differentiation
Let $y = x^{(x^x)}$. $\ln y = x^x \ln x$.
Step 2: Differentiate
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x) \cdot \ln x + x^x \cdot \frac{1}{x}$.
Note: $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$.
Step 3: Calculation
$\frac{1}{y} \frac{dy}{dx} = x^x(1 + \ln x) \ln x + x^{x-1}$.
$\frac{dy}{dx} = x^{(x^x)} [x^{x-1} + x^x \ln x (1 + \ln x)]$.
Final Answer: (D)
Let $y = x^{(x^x)}$. $\ln y = x^x \ln x$.
Step 2: Differentiate
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x) \cdot \ln x + x^x \cdot \frac{1}{x}$.
Note: $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$.
Step 3: Calculation
$\frac{1}{y} \frac{dy}{dx} = x^x(1 + \ln x) \ln x + x^{x-1}$.
$\frac{dy}{dx} = x^{(x^x)} [x^{x-1} + x^x \ln x (1 + \ln x)]$.
Final Answer: (D)
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