if y=[log{x+√(1+x^2)}]^2 then find the value of nth derivative of y at x=0

To find the value of the nth derivative of the function \( y = [\log(x + \sqrt{1 + x^2})]^2 \) at \( x = 0 \), we can start by simplifying the expression for \( y \). This function is related to hyperbolic functions, which can make the differentiation process more manageable.

Understanding the Function

The expression inside the logarithm can be rewritten using hyperbolic identities. Notice that:

• We can express \( x + \sqrt{1 + x^2} \) as \( e^{\text{arsinh}(x)} \), where \( \text{arsinh}(x) \) is the inverse hyperbolic sine function.
• This leads to \( \log(x + \sqrt{1 + x^2}) = \text{arsinh}(x) \).

Thus, we can rewrite \( y \) as:

\( y = [\text{arsinh}(x)]^2 \).

Finding the Derivatives

Next, we need to find the derivatives of \( y \) at \( x = 0 \). The Taylor series expansion of \( \text{arsinh}(x) \) around \( x = 0 \) is:

\( \text{arsinh}(x) = x + \frac{x^3}{6} + \frac{x^5}{40} + O(x^7) \).

Squaring this series gives us:

\( y = \left( x + \frac{x^3}{6} + \frac{x^5}{40} + O(x^7) \right)^2 \).

Expanding this, we have:

• \( y = x^2 + 2 \cdot x \cdot \frac{x^3}{6} + \frac{x^6}{36} + O(x^8) \)
• \( = x^2 + \frac{x^4}{3} + O(x^6) \).

Evaluating the nth Derivative

To find the nth derivative of \( y \) at \( x = 0 \), we observe the series expansion. The only non-zero derivatives at \( x = 0 \) will be for even powers of \( x \) since all terms in the expansion are even. Specifically:

• The coefficient of \( x^2 \) contributes to the second derivative.
• The coefficient of \( x^4 \) contributes to the fourth derivative.

For \( n = 2 \), the second derivative \( y''(0) = 2! \cdot 1 = 2 \). For \( n = 4 \), the fourth derivative \( y^{(4)}(0) = 4! \cdot \frac{1}{3} = \frac{24}{3} = 8 \). For any odd \( n \), the derivative will be zero since there are no odd powers in the expansion.

Final Result

In summary, the value of the nth derivative of \( y \) at \( x = 0 \) can be expressed as:

• If \( n = 2 \), then \( y^{(2)}(0) = 2 \).
• If \( n = 4 \), then \( y^{(4)}(0) = 8 \).
• If \( n \) is odd, \( y^{(n)}(0) = 0 \).

This pattern continues for higher even derivatives, where you can compute them similarly based on the coefficients of the series expansion. Thus, the nth derivative at \( x = 0 \) depends on whether \( n \) is even or odd, and the specific values can be calculated accordingly.