1. **Problem Statement:** State Leibnitz theorem for the nth derivative of the product of two functions and find the nth derivative of (i) $x^3 \sin x$ and (ii) $x^2 e^{2x}$. 2. **Leibnitz Theorem:** For two functions $u(x)$ and $v(x)$, the nth derivative of their product is given by: $$\frac{d^n}{dx^n}[u(x)v(x)] = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{dx^{n-k}} \frac{d^k v}{dx^k}$$ This means the nth derivative of the product is the sum of products of derivatives of $u$ and $v$ whose orders add up to $n$. 3. **Find $\frac{d^n}{dx^n} (x^3 \sin x)$:** Let $u = x^3$, $v = \sin x$. - Derivatives of $u$: $u^{(m)} = \frac{d^m}{dx^m} x^3$. - $u^{(0)} = x^3$ - $u^{(1)} = 3x^2$ - $u^{(2)} = 6x$ - $u^{(3)} = 6$ - $u^{(m)} = 0$ for $m > 3$ - Derivatives of $v$: $v^{(k)} = \sin^{(k)} x$ cycles every 4 derivatives: - $v^{(0)} = \sin x$ - $v^{(1)} = \cos x$ - $v^{(2)} = -\sin x$ - $v^{(3)} = -\cos x$ - $v^{(4)} = \sin x$ and so on. Using Leibnitz theorem: $$\frac{d^n}{dx^n} (x^3 \sin x) = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)}$$ Since $u^{(m)}=0$ for $m>3$, terms with $n-k > 3$ vanish, so $k \ge n-3$. Explicitly: $$\frac{d^n}{dx^n} (x^3 \sin x) = \sum_{k=\max(0, n-3)}^n \binom{n}{k} u^{(n-k)} v^{(k)}$$ 4. **Find $\frac{d^n}{dx^n} (x^2 e^{2x})$:** Let $u = x^2$, $v = e^{2x}$. - Derivatives of $u$: - $u^{(0)} = x^2$ - $u^{(1)} = 2x$ - $u^{(2)} = 2$ - $u^{(m)} = 0$ for $m > 2$ - Derivatives of $v$: - $v^{(k)} = (2)^k e^{2x}$ Using Leibnitz theorem: $$\frac{d^n}{dx^n} (x^2 e^{2x}) = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)}$$ Only terms with $n-k \le 2$ survive, so $k \ge n-2$. Explicitly: $$\frac{d^n}{dx^n} (x^2 e^{2x}) = \sum_{k=\max(0, n-2)}^n \binom{n}{k} u^{(n-k)} (2)^k e^{2x}$$ 5. **Summary:** - Leibnitz theorem formula is the key. - For polynomials times other functions, derivatives of polynomial vanish after finite order. - Use the finite number of terms accordingly. Final answers: $$\boxed{\frac{d^n}{dx^n} (x^3 \sin x) = \sum_{k=\max(0, n-3)}^n \binom{n}{k} u^{(n-k)} v^{(k)}}$$ where $u^{(m)}$ are derivatives of $x^3$ and $v^{(k)}$ are derivatives of $\sin x$. $$\boxed{\frac{d^n}{dx^n} (x^2 e^{2x}) = \sum_{k=\max(0, n-2)}^n \binom{n}{k} u^{(n-k)} (2)^k e^{2x}}$$ where $u^{(m)}$ are derivatives of $x^2$.