The question asks to compute the third derivative of $y = e^{-2x} \ln x$ using Leibniz's Theorem. Here's the detailed solution:

Leibniz's Theorem

For two functions $u(x)$ and $v(x)$, the $n$-th derivative of their product is given by: $\frac{d^n}{dx^n}[u(x)v(x)] = \sum_{k=0}^n \binom{n}{k} u^{(k)}(x) v^{(n-k)}(x)$

Apply this theorem:

Here $y = e^{-2x} \ln x$, so:

• $u(x) = e^{-2x}$
• $v(x) = \ln x$

We need the third derivative: $\frac{d^3y}{dx^3}$.

Step 1: Derivatives of $u(x) = e^{-2x}$

1. $u(x) = e^{-2x}$
2. First derivative: $u'(x) = -2e^{-2x}$
3. Second derivative: $u''(x) = 4e^{-2x}$
4. Third derivative: $u'''(x) = -8e^{-2x}$

Step 2: Derivatives of $v(x) = \ln x$

1. $v(x) = \ln x$
2. First derivative: $v'(x) = \frac{1}{x}$
3. Second derivative: $v''(x) = -\frac{1}{x^2}$
4. Third derivative: $v'''(x) = \frac{2}{x^3}$

Step 3: Apply Leibniz's Theorem for $n = 3$

Using: $\frac{d^3}{dx^3} [u(x) v(x)] = \sum_{k=0}^3 \binom{3}{k} u^{(k)}(x) v^{(3-k)}(x)$

$\frac{d^3y}{dx^3} = u^{(0)}(x) v^{(3)}(x) + 3 u^{(1)}(x) v^{(2)}(x) + 3 u^{(2)}(x) v^{(1)}(x) + u^{(3)}(x) v^{(0)}(x)$

Substitute derivatives:

$\frac{d^3y}{dx^3} = e^{-2x} \cdot \frac{2}{x^3} + 3(-2e^{-2x}) \cdot \left(-\frac{1}{x^2}\right) + 3(4e^{-2x}) \cdot \frac{1}{x} + (-8e^{-2x}) \cdot \ln x$

Simplify each term:

1. $e^{-2x} \cdot \frac{2}{x^3} = \frac{2e^{-2x}}{x^3}$
2. $3(-2e^{-2x}) \cdot \left(-\frac{1}{x^2}\right) = \frac{6e^{-2x}}{x^2}$
3. $3(4e^{-2x}) \cdot \frac{1}{x} = \frac{12e^{-2x}}{x}$
4. $(-8e^{-2x}) \cdot \ln x = -8e^{-2x} \ln x$

Combine terms: $\frac{d^3y}{dx^3} = \frac{2e^{-2x}}{x^3} + \frac{6e^{-2x}}{x^2} + \frac{12e^{-2x}}{x} - 8e^{-2x} \ln x$

Final Answer:

$\frac{d^3y}{dx^3} = e^{-2x} \left( \frac{2}{x^3} + \frac{6}{x^2} + \frac{12}{x} - 8\ln x \right)$

Let me know if you'd like any part of this explained further or expanded!

Related Questions:

1. How can Leibniz's theorem be generalized to higher derivatives?
2. What are the practical applications of using $e^{-2x} \ln x$ in differential equations?
3. How does Leibniz's theorem simplify calculations compared to direct differentiation?
4. Can you solve similar problems where $y$ involves trigonometric or exponential products?
5. How does the behavior of $y$ change for very large or very small $x$?

Tip: Always calculate derivatives of simpler terms like $u(x)$ and $v(x)$ beforehand to reduce errors in complex applications like Leibniz's theorem.