Leibnitz Theorem
This theorem is useful in finding nth derivative of product of two functions.
If \(u,v\) are two functions of \(x\), then
\[ D^n(uv) = D^n(u)v \hspace{5pt} + \hspace{5pt} ^n\hspace{-3pt}C_1D^{n-1}(u)D(v) \]
\[ +\hspace{5pt} ^n\hspace{-4pt}C_2D^{n-2}(u)D^2(v) \hspace{5pt} + ...... + ^n\hspace{-4pt}C_n(u)D^n(v) \]
Sample Problem
Find nth derivative of \(x^3\cos{x} \)
Solution
One thing that we will have to observe is to take which function as \(u\) \(v\), because if you observe the formula carefully, its long ofcourse, but it can be shortened just by choosing the function which reduces to 0 after 2 or 3 or 4 derivatives. In this example, \(x^3\) will become 0 after 3 derivatives only so this will be \(v\) and \(\cos{x} \) will be \(u\). \[ D^n(x^3\cos{x}) = D^n(\cos{x})x^3 + ^n\hspace{-3pt}C_1 D^{n-1}(\cos{x})D(x^3) \] \[ + ^n\hspace{-3pt}C_2 D^{n-2}(\cos{x})D^2(x^3) + ^n\hspace{-3pt}C_3D^{n-3}(\cos{x})D^3(x^3) \] \[ + ^n\hspace{-3pt}C_4 D^{n-4}(\cos{x})D^4(x^3) \] \[ = x^3\cos{\left(x + \frac{n\pi}{2} \right)} + n(3x^2)\cos{\left[x + \frac{(n-1)\pi}{2} \right]} \] \[ + \frac{n(n-1)}{1\times 2}\cos{\left[x + \frac{(n-2)\pi}{2} \right]}(6x) \] \[ + \frac{n(n-1)(n-2)}{1\times 2\times 3}\cos{\left[x + \frac{(n-3)\pi}{2} \right]}(6) \] \[ = x^3\cos{\left(x + \frac{n\pi}{2} \right)} + 3x^2n\sin{\left(x + \frac{n\pi}{2} \right)} \] \[ - 3n(n-1)x\cos{\left(x + \frac{n\pi}{2} \right)} \] \[ - n(n-1)(n-2)\sin{\left(x + \frac{n\pi}{2} \right)} \] \[ \Rightarrow D^n(x^3\cos{x}) = x\cos{(x+\frac{n\pi}{2})}\left[x^2 - 3n(n-1) \right] \] \[ + n\sin{\left(x + \frac{n\pi}{2} \right)}\left[3x^2 - (n-1)(n-2) \right] \]
Sample Problem 2
If \( y = \left(\sin^{-1}{x} \right)^2 \) prove that \[(1 - x^2)y_{n+2} - (2n+1)xy_{n=1} - n^2y_n = 0 \]
Solution
\[ y = \left(\sin^{-1}{x} \right)^2 \] \[ y_1 = 2(\sin^{-1}{x})\times \frac{1}{\sqrt{1 - x^2}} \] \[ \sqrt{1 - x^2}y_1 = 2(\sin^{-1}{x}) \] Squaring on both sides \[ (1 - x^2)y^2_1 = 4(\sin^{-1}{x})^2 \] \[ (1 - x^2)y_1 = 4y \] Differentiating on both sides \[ (1 - x^2)2y_1y_2 + (-2x)y_1^2 = 4y_1 \] \[ (1 - x^2)2y_1y_2 - 2xy_1^2 - 4y_1 = 0 \] \[ 2y_1 \left[(1-x^2)y_2 - xy_1 - 2 \right] = 0 \] \[ \left[(1-x^2)y_2 - xy_1 - 2 \right] = 0 \] Now differentiating nth times by Leibnitz Theorem \[ D^n\left[(1-x^2)y_2 \right] - D^n(xy_1) - D^n(2) = 0 \] \[ \left[D_n(y_2)(1-x^2) + ^n\hspace{-3pt}C_1 D^{n-1}(y_2)D(1-x^2) + ^n\hspace{-3pt}C_2 D^{n-2}(y_2)D^2(1-x^2) \right] \] \[ - \left[D^n(y_1)x + ^n\hspace{-3pt}C_1 D^{n-1}(y_1)D(x) \right] = 0 \] \[ (1-x^2)y_{n+2} + ny_{n+1}(-2x) + \frac{n(n-1)}{1\times 2}y_n (-2) \] \[ - \left[xy_{n+1} + ny_n \right] = 0 \] \[ (1-x^2)y_{n+2} - 2xny_{n+1} - (n^2 - n)y_n \] \[ - xy^{n+1} - ny_n \] \[ (1-x^2)y_{n+2} - 2xny_{n+1} - n^2y_n + \cancel{ny_n} \] \[ - xy_{n+1} - \cancel{ny_n} = 0 \] \[ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2y_n = 0 \] Hence Proved :)
