If `y= (x^x )^x ` then `(dy)/(dx)` =

Options

A. `xy (1 + log x)`
B. `XY (1+2 log x)`
C. ` x/y (1 + log X)`
D. `(x)/(y ) (1+ 2 log x)`

Answer

To find the derivative of the function \( y = (x^x)^x \), we can follow these steps:

### Step 1: Simplify the expression
First, we simplify the expression \( y = (x^x)^x \) using the property of exponents:
\[ y = x^{x \cdot x} = x^{x^2} \]

### Step 2: Take the natural logarithm
Next, we take the natural logarithm of both sides to facilitate differentiation:
\[ \ln y = \ln(x^{x^2}) \]

### Step 3: Apply the logarithmic property
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the right side:
\[ \ln y = x^2 \ln x \]

### Step 4: Differentiate both sides
Now we differentiate both sides with respect to \( x \). Remember to use implicit differentiation on the left side:
\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we apply the product rule:
\[ \frac{d}{dx}(x^2 \ln x) = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x) \]
Calculating the derivatives:
\[ \frac{d}{dx}(x^2) = 2x \quad \text{and} \quad \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Thus, we have:
\[ \frac{d}{dx}(x^2 \ln x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x \]

### Step 5: Set the derivatives equal
Now we set the derivatives equal to each other:
\[ \frac{1}{y} \frac{dy}{dx} = 2x \ln x + x \]

### Step 6: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y(2x \ln x + x) \]

### Step 7: Substitute back for \( y \)
Recall that \( y = x^{x^2} \):
\[ \frac{dy}{dx} = x^{x^2}(2x \ln x + x) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = x^{x^2} (2x \ln x + x) \]

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