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Exercise 9 Use Leibnitz Theorem to find the

\mathrm{n}^{\text {th }}

order de...

Jun 18, 2024

Solution by Steps

step 1

To find the

n^{\text{th}}

order derivative of

y = x^3 \sin(2x)

using Leibniz's theorem, we start by expressing the function in a form suitable for the theorem

step 2

Leibniz's theorem for the

n^{\text{th}}

derivative of a product

u(x)v(x)

is given by:

\frac{d^n}{dx^n} [u(x)v(x)] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} [u(x)] \cdot \frac{d^k}{dx^k} [v(x)]

Here,

u(x) = x^3

and

v(x) = \sin(2x)

step 3

Calculate the derivatives of

u(x) = x^3

\frac{d^m}{dx^m} [x^3] = \begin{cases} 0 &amp; \text{if } m &gt; 3 \\ 6x &amp; \text{if } m = 2 \\ 3x^2 &amp; \text{if } m = 1 \\ x^3 &amp; \text{if } m = 0 \end{cases}

step 4

Calculate the derivatives of

v(x) = \sin(2x)

\frac{d^k}{dx^k} [\sin(2x)] = 2^k \sin\left(2x + \frac{k\pi}{2}\right)

step 5

Substitute these derivatives into Leibniz's theorem:

\frac{d^n}{dx^n} [x^3 \sin(2x)] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} [x^3] \cdot \frac{d^k}{dx^k} [\sin(2x)]

step 6

Simplify the expression by evaluating the sum for specific values of

k

and

n

Answer

The

n^{\text{th}}

order derivative of

y = x^3 \sin(2x)

using Leibniz's theorem is given by the sum:

\frac{d^n}{dx^n} [x^3 \sin(2x)] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} [x^3] \cdot \frac{d^k}{dx^k} [\sin(2x)]

Key Concept

Leibniz's Theorem

Explanation

Leibniz's theorem allows us to find the

n^{\text{th}}

derivative of a product of two functions by expressing it as a sum of products of the derivatives of the individual functions.