Leibniz's Theorem in Calculus

Leibniz's Theorem

• Leibniz's theorem provides a formula for the nth derivative of a product of two functions.
• It's a powerful tool for finding higher-order derivatives of complicated functions without repeated application of the product rule.
• This theorem significantly simplifies the process of finding higher-order derivatives of product functions compared with repeated applications of the standard product rule.

Formula

• If u(x) and v(x) are functions of x differentiable n times, the nth derivative of their product is given by:

  dⁿ/dxⁿ (u(x)v(x)) = Σ_k=0ⁿ (n choose k) * (d^ku/dx^k)(d^(n-k)v/dx^(n-k))

• "(n choose k)" denotes the binomial coefficient, calculated as n! / (k! * (n-k)!).

Example

• Finding the 3rd derivative of f(x) = x²sin(x):

• Let u(x) = x² and v(x) = sin(x).

• Calculate the first three derivatives of u(x) and v(x): u'(x) = 2x, u''(x) = 2, u'''(x) = 0 v'(x) = cos(x), v''(x) = -sin(x), v'''(x) = -cos(x)

• Using Leibniz's theorem:

f'''(x) = (3 choose 0) * u'''(x) * v(x) + (3 choose 1) * u''(x) * v'(x) + (3 choose 2) * u'(x) * v''(x) + (3 choose 3) * u(x) * v'''(x) = 1 * 0 * sin(x) + 3 * 2 * cos(x) + 3 * 2x * (-sin(x)) + 1 * x² * (-cos(x)) = 6cos(x) - 6xsin(x) - x²cos(x)

Applications

• Leibniz's theorem is used extensively in various fields of engineering and mathematics.
• It is crucial for solving differential equations, especially those containing products of functions.
• It simplifies higher-order derivative calculations, enhancing computational efficiency.

Significance

• This theorem generalizes the product rule for differentiation, making calculating higher-order derivatives of product functions more manageable.
• This simplification significantly streamlines the process of obtaining solutions to differential equations involving products of functions.
• It drastically reduces calculation steps compared to applying the product rule repeatedly.

This quiz explores Leibniz's theorem, a key concept in calculus that details the nth derivative of a product of two functions. Test your understanding of the formula and its applications through examples, including higher-order derivatives. Get ready to deepen your calculus skills!