If `y=x^(x^(2)),` then `(dy)/(dx)` equals

Options

A. `2logx.x^(2)`
B. `(2logx+1).x^(x^(2))`
C. `(2logx+1).x^(x^(2))+1`
D. `x^(x^(2)+1).(log(ex^(2)))`

Explanation

To find the derivative of the function \( y = x^{x^2} \), we can follow these steps:

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

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

### Step 3: Differentiate both sides
Next, we differentiate both sides with respect to \( x \). Remember that when differentiating \( \log y \), we will use implicit differentiation:
\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]
For the right-hand side, we will use the product rule since we have \( x^2 \log x \):
\[ \frac{d}{dx}(x^2 \log x) = \frac{d}{dx}(x^2) \cdot \log x + x^2 \cdot \frac{d}{dx}(\log x) \]
Calculating these derivatives:
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( \log x \) is \( \frac{1}{x} \).

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

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

### Step 5: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y(2x \log x + x) \]

### Step 6: Substitute back for \( y \)
Recall that \( y = x^{x^2} \):
\[ \frac{dy}{dx} = x^{x^2}(2x \log x + x) \]

### Step 7: Simplify the expression
We can factor out \( x \) from the expression:
\[ \frac{dy}{dx} = x^{x^2} \cdot x (2 \log x + 1) = x^{x^2 + 1}(2 \log x + 1) \]

Thus, the final result is:
\[ \frac{dy}{dx} = x^{x^2 + 1}(2 \log x + 1) \]