Find nth derivative of the following question xlogx by leibniz's theorem

To find the nth derivative of the function \( f(x) = x \log x \) using Leibniz's theorem, we first need to understand what Leibniz's theorem entails. This theorem provides a formula for the nth derivative of a product of two functions. In this case, we can treat \( f(x) \) as the product of two functions: \( u(x) = x \) and \( v(x) = \log x \).

Applying Leibniz's Theorem

Leibniz's theorem states that if you have two functions \( u(x) \) and \( v(x) \), the nth derivative of their product is given by:

(uv)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(k)}(x) v^{(n-k)}(x)

Here, \( u^{(k)}(x) \) is the k-th derivative of \( u(x) \) and \( v^{(n-k)}(x) \) is the (n-k)-th derivative of \( v(x) \). Now, let’s compute the derivatives of \( u(x) \) and \( v(x) \).

Finding Derivatives of u(x) and v(x)

• For \( u(x) = x \):
  • First derivative: \( u'(x) = 1 \)
  • Second derivative: \( u''(x) = 0 \)
  • All higher derivatives: \( u^{(k)}(x) = 0 \) for \( k \geq 2 \)
• For \( v(x) = \log x \):
  • First derivative: \( v'(x) = \frac{1}{x} \)
  • Second derivative: \( v''(x) = -\frac{1}{x^2} \)
  • Third derivative: \( v'''(x) = \frac{2}{x^3} \)
  • In general, \( v^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{x^n} \) for \( n \geq 1 \)

Calculating the nth Derivative

Now that we have the derivatives of both functions, we can substitute them into Leibniz's formula. Since \( u^{(k)}(x) = 0 \) for \( k \geq 2 \), the only non-zero contributions to the sum will come from \( k = 0 \) and \( k = 1 \).

Thus, we can simplify the expression:

(x \log x)^{(n)} = u^{(0)}(x) v^{(n)}(x) + u^{(1)}(x) v^{(n-1)}(x)

Substituting the derivatives we found:

(x \log x)^{(n)} = x \cdot \left((-1)^{n-1} \frac{(n-1)!}{x^n}\right) + 1 \cdot \left((-1)^{n-2} \frac{(n-2)!}{x^{n-1}}\right)

Final Expression

This leads us to:

(x \log x)^{(n)} = (-1)^{n-1} \frac{(n-1)!}{x^{n-1}} + (-1)^{n-2} \frac{(n-2)!}{x^{n-1}}

Combining the terms gives us:

(x \log x)^{(n)} = \frac{(-1)^{n-2}}{x^{n-1}} \left((n-1)! - (n-2)!\right)

For \( n \geq 2 \), this can be further simplified to:

(x \log x)^{(n)} = \frac{(-1)^{n-2} (n-2)! (n-1)}{x^{n-1}}

Summary

In summary, using Leibniz's theorem, we derived the nth derivative of the function \( x \log x \). The key steps involved calculating the derivatives of the individual components and applying the theorem correctly. This method not only provides a systematic approach to finding derivatives of products but also deepens our understanding of how derivatives interact in more complex functions.