Differentiate the following w.r.t. x :
`x^(x^(2))`

Answer

To differentiate the function \( y = x^{x^2} \) with respect to \( x \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln(x^{x^2}) \]
Using the logarithmic property, we can simplify the right-hand side:
\[ \ln y = x^2 \ln x \]

### Step 2: Differentiate both sides with respect to \( x \)
Now, we differentiate both sides with respect to \( x \). Remember to use implicit differentiation on the left side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x^2 \ln x) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2 \ln x) \]

### Step 3: Apply the product rule on the right side
For the right-hand 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:
\[ \frac{d}{dx}(x^2) = 2x \quad \text{and} \quad \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Substituting these back into the equation gives:
\[ \frac{d}{dx}(x^2 \ln x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x \]

### Step 4: Substitute back into the equation
Now we can substitute this result back into our differentiation equation:
\[ \frac{1}{y} \frac{dy}{dx} = 2x \ln x + x \]

### Step 5: 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 6: Substitute back for \( y \)
Since \( y = x^{x^2} \), we substitute back:
\[ \frac{dy}{dx} = x^{x^2}(2x \ln x + x) \]

### Final Answer
Thus, the derivative of \( y = x^{x^2} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{x^2}(2x \ln x + x) \]

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