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

Answer

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

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln(x^{x^2}) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the equation:
\[ \ln y = x^2 \ln x \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. For the left side, we use implicit differentiation:
\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we will use 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 these derivatives:
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \).

So, we have:
\[ \frac{d}{dx}(x^2 \ln x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x \]

### Step 4: Set the derivatives equal
Now we set the derivatives from both sides equal to each other:
\[ \frac{1}{y} \frac{dy}{dx} = 2x \ln x + x \]

### 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 we defined \( y = x^{x^2} \), 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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