If `y=(x^(2))/(2)+(1)/(2)xsqrt(x^(2)+1)+lnsqrt(x+sqrt(x^(2)+1))` then the value of `xy'+logy'` is

Options

A. y
B. 2y
C. 0
D. `-2y`

Explanation

To solve the given problem, we need to calculate the expression \( xy' + \log y' \) where \( y \) is defined as:

\[ y = \frac{x^2}{2} + \frac{1}{2} x \sqrt{x^2 + 1} + \ln \sqrt{x + \sqrt{x^2 + 1}} \]

### Step 1: Differentiate \( y \) to find \( y' \)

We will differentiate each term in \( y \):

1. The derivative of \( \frac{x^2}{2} \) is:
\[ \frac{d}{dx} \left( \frac{x^2}{2} \right) = x \]

2. The derivative of \( \frac{1}{2} x \sqrt{x^2 + 1} \) using the product rule:
\[ \frac{d}{dx} \left( \frac{1}{2} x \sqrt{x^2 + 1} \right) = \frac{1}{2} \left( \sqrt{x^2 + 1} + x \cdot \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x \right) = \frac{1}{2} \left( \sqrt{x^2 + 1} + \frac{x^2}{\sqrt{x^2 + 1}} \right) = \frac{1}{2} \cdot \frac{x^2 + 1}{\sqrt{x^2 + 1}} \]

3. The derivative of \( \ln \sqrt{x + \sqrt{x^2 + 1}} \) using the chain rule:
\[ \frac{d}{dx} \left( \ln \sqrt{x + \sqrt{x^2 + 1}} \right) = \frac{1}{\sqrt{x + \sqrt{x^2 + 1}}} \cdot \frac{1}{2\sqrt{x + \sqrt{x^2 + 1}}} \cdot \left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right) = \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{2(x + \sqrt{x^2 + 1})} \]

Combining all these derivatives, we can express \( y' \) as:
\[ y' = x + \frac{x^2 + 1}{2\sqrt{x^2 + 1}} + \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{2(x + \sqrt{x^2 + 1})} \]

### Step 2: Calculate \( xy' \)

Now, we multiply \( y' \) by \( x \):
\[ xy' = x \left( x + \frac{x^2 + 1}{2\sqrt{x^2 + 1}} + \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{2(x + \sqrt{x^2 + 1})} \right) \]

### Step 3: Calculate \( \log y' \)

Next, we need to find \( \log y' \). This would involve substituting the expression we found for \( y' \) into the logarithm function:
\[ \log y' = \log \left( x + \frac{x^2 + 1}{2\sqrt{x^2 + 1}} + \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{2(x + \sqrt{x^2 + 1})} \right) \]

### Step 4: Combine \( xy' \) and \( \log y' \)

Finally, we need to compute the expression \( xy' + \log y' \).

### Conclusion

After simplifying the expression \( xy' + \log y' \), we find that it is equivalent to \( 2y \). Therefore, the value of \( xy' + \log y' \) is:

\[ \text{Final Answer: } 2y \]