find the nth derivative ofx^2e^(3x) sin4x using leibniz theorem

To find the nth derivative of the function \( f(x) = x^2 e^{3x} \sin(4x) \) using Leibniz's theorem, we first need to understand how this theorem applies to products of functions. Leibniz's theorem states that the nth derivative of a product of two functions can be expressed as a sum involving the derivatives of each function. Let's break this down step by step.

Understanding Leibniz's Theorem

Leibniz's theorem can be stated as follows: if you have two functions \( u(x) \) and \( v(x) \), the nth derivative of their product is given by:

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

Here, \( u^{(k)} \) is the k-th derivative of \( u \), and \( v^{(n-k)} \) is the (n-k)-th derivative of \( v \). In our case, we can identify:

• \( u(x) = x^2 \)
• \( v(x) = e^{3x} \sin(4x) \)

Finding Derivatives of Each Function

Next, we need to compute the derivatives of \( u(x) \) and \( v(x) \). The derivatives of \( u(x) = x^2 \) are straightforward:

• \( u^{(0)} = x^2 \)
• \( u^{(1)} = 2x \)
• \( u^{(2)} = 2 \)
• \( u^{(k)} = 0 \) for \( k \geq 3 \)

Now, let's focus on \( v(x) = e^{3x} \sin(4x) \). To find its derivatives, we can use the product rule and the chain rule. The derivatives of \( v(x) \) can be computed using the following identities:

• The derivative of \( e^{ax} \) is \( ae^{ax} \).
• The derivative of \( \sin(bx) \) is \( b\cos(bx) \).

Using these rules, we can derive a pattern for \( v^{(n)} \). The nth derivative of \( v(x) \) can be expressed as a linear combination of \( e^{3x} \sin(4x) \) and \( e^{3x} \cos(4x) \). Specifically, we can use the formula:

\( v^{(n)} = e^{3x} \left( A_n \sin(4x) + B_n \cos(4x) \right) \)

where \( A_n \) and \( B_n \) are coefficients that depend on \( n \). These coefficients can be found using recurrence relations derived from the derivatives of sine and cosine functions.

Combining the Results

Now that we have the derivatives of both \( u \) and \( v \), we can apply Leibniz's theorem:

\( (x^2 e^{3x} \sin(4x))^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(k)} v^{(n-k)} \)

Substituting the derivatives we found:

• For \( k = 0 \): \( u^{(0)} = x^2 \) and \( v^{(n)} \)
• For \( k = 1 \): \( u^{(1)} = 2x \) and \( v^{(n-1)} \)
• For \( k = 2 \): \( u^{(2)} = 2 \) and \( v^{(n-2)} \)

For \( k \geq 3 \), \( u^{(k)} = 0 \), so those terms do not contribute to the sum. Thus, we can express the nth derivative as:

\( (x^2 e^{3x} \sin(4x))^{(n)} = \binom{n}{0} x^2 v^{(n)} + \binom{n}{1} 2x v^{(n-1)} + \binom{n}{2} 2 v^{(n-2)} \)

Final Expression

In conclusion, the nth derivative of \( x^2 e^{3x} \sin(4x) \) can be computed using the formula derived above, where \( v^{(n-k)} \) can be expressed in terms of \( e^{3x} \sin(4x) \) and \( e^{3x} \cos(4x) \) with the coefficients \( A_n \) and \( B_n \) determined by the order of the derivative. This approach not only simplifies the calculation but also provides a structured way to analyze the behavior of the function as we take higher derivatives.