If ` y=x^(x) ,then xy (d^(2)y)/(dx^(2))-x((dy)/(dx) )^(2)=`

Options

A. ` -y^(2)`
B. ` y^(2)`
C. ` -xy^(2)`
D. `xy^(2)`

Answer

To solve the problem, we need to find the expression \( xy \frac{d^2y}{dx^2} - x \left( \frac{dy}{dx} \right)^2 \) given that \( y = x^x \).

### Step-by-Step Solution:

1. **Define the function**:
\[ y = x^x \]

2. **Take the logarithm of both sides**:
\[ \log y = \log(x^x) = x \log x \]

3. **Differentiate both sides with respect to \( x \)**:
Using implicit differentiation:
\[ \frac{1}{y} \frac{dy}{dx} = \log x + 1 \]
Therefore,
\[ \frac{dy}{dx} = y(\log x + 1) \]

4. **Substitute \( y \) back**:
Since \( y = x^x \):
\[ \frac{dy}{dx} = x^x (\log x + 1) \]

5. **Differentiate again to find \( \frac{d^2y}{dx^2} \)**:
We apply the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( x^x (\log x + 1) \right) \]
Let \( u = x^x \) and \( v = \log x + 1 \):
\[ \frac{d^2y}{dx^2} = \frac{du}{dx} v + u \frac{dv}{dx} \]

- First, find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = x^x (\log x + 1) \]
- Now find \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{1}{x} \]

Putting it all together:
\[ \frac{d^2y}{dx^2} = x^x (\log x + 1)(\log x + 1) + x^x \cdot \frac{1}{x} \]
Simplifying:
\[ \frac{d^2y}{dx^2} = x^x (\log x + 1)^2 + x^{x-1} \]

6. **Substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the expression**:
Now we need to compute:
\[ xy \frac{d^2y}{dx^2} - x \left( \frac{dy}{dx} \right)^2 \]
Substitute \( y = x^x \):
\[ xy \frac{d^2y}{dx^2} = x \cdot x^x \left( x^x (\log x + 1)^2 + x^{x-1} \right) \]
\[ = x^{x+1} \left( x^x (\log x + 1)^2 + x^{x-1} \right) \]

Now for \( x \left( \frac{dy}{dx} \right)^2 \):
\[ x \left( \frac{dy}{dx} \right)^2 = x \left( x^x (\log x + 1) \right)^2 = x \cdot x^{2x} (\log x + 1)^2 = x^{2x + 1} (\log x + 1)^2 \]

7. **Combine the two parts**:
\[ xy \frac{d^2y}{dx^2} - x \left( \frac{dy}{dx} \right)^2 = x^{x+1} \left( x^x (\log x + 1)^2 + x^{x-1} \right) - x^{2x + 1} (\log x + 1)^2 \]

Simplifying:
\[ = x^{x+1} x^{x-1} + x^{x+1} x^x (\log x + 1)^2 - x^{2x + 1} (\log x + 1)^2 \]
\[ = x^{2x} + x^{2x + 1}(\log x + 1)^2 - x^{2x + 1}(\log x + 1)^2 \]
\[ = x^{2x} \]

8. **Final Result**:
\[ xy \frac{d^2y}{dx^2} - x \left( \frac{dy}{dx} \right)^2 = x^{2x} \]