Leibnitz Theorem Engineering Maths, Btech first year

Leibnitz Theorem

This theorem is useful in finding nth derivative of product of two functions.
If $u,v$ are two functions of $x$, then $$D^n(uv)=D^n(u)v+{}^n{C}_1{D}^{n-1}(u)D(v)$$ $$+{}^n{C}_2{D}^{n-2}(u)D^2(v)+......{+}^n{C}_n(u)D^n(v)$$

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 $u$. $$D^n(x^3\cos x)=D^n(\cos x)x^3{+}^n{C}_1{D}^{n-1}(\cos x)D(x^3)$$ $${+}^n{C}_2{D}^{n-2}(\cos x)D^2(x^3){+}^n{C}_3{D}^{n-3}(\cos x)D^3(x^3)$$ $${+}^n{C}_4{D}^{n-4}(\cos x)D^4(x^3)$$ $$=x^3\cos (x+\frac{n\pi }{2})+n(3x^2)\cos [x+\frac{(n-1)\pi }{2}]$$ $$+\frac{n(n-1)}{1\times 2}\cos [x+\frac{(n-2)\pi }{2}](6x)$$ $$+\frac{n(n-1)(n-2)}{1\times 2\times 3}\cos [x+\frac{(n-3)\pi }{2}](6)$$ $$=x^3\cos (x+\frac{n\pi }{2})+3x^{2}n\sin (x+\frac{n\pi }{2})$$ $$-3n(n-1)x\cos (x+\frac{n\pi }{2})$$ $$-n(n-1)(n-2)\sin (x+\frac{n\pi }{2})$$ $$\Rightarrow D^n(x^3\cos x)=x\cos (x+\frac{n\pi }{2})[x^2-3n(n-1)]$$ $$+n\sin (x+\frac{n\pi }{2})[3x^2-(n-1)(n-2)]$$

Sample Problem 2

If $y={({\sin }^{-1}x)}^2$ prove that $$(1-x^2)y_{n+2}-(2n+1)x{y}_{n=1}-n^2{y}_n=0$$

Solution

$$y={({\sin }^{-1}x)}^2$$ $$y_1=2({\sin }^{-1}x)\times \frac{1}{\sqrt{1-x^2}}$$ $$\sqrt{1-x^2}{y}_1=2({\sin }^{-1}x)$$ Squaring on both sides $$(1-x^2)y_1^2=4({\sin }^{-1}x{)}^2$$ $$(1-x^2)y_1=4y$$ Differentiating on both sides $$(1-x^2)2y_1{y}_2+(-2x)y_1^2=4y_1$$ $$(1-x^2)2y_1{y}_2-2x{y}_1^2-4y_1=0$$ $$2y_1[(1-x^2)y_2-x{y}_1-2]=0$$ $$[(1-x^2)y_2-x{y}_1-2]=0$$ Now differentiating nth times by Leibnitz Theorem $$D^n[(1-x^2)y_2]-D^n(x{y}_1)-D^n(2)=0$$ $$[D_n(y_2)(1-x^2){+}^n{C}_1{D}^{n-1}(y_2)D(1-x^2){+}^n{C}_2{D}^{n-2}(y_2)D^2(1-x^2)]$$ $$-[D^n(y_1)x{+}^n{C}_1{D}^{n-1}(y_1)D(x)]=0$$ $$(1-x^2)y_{n+2}+n{y}_{n+1}(-2x)+\frac{n(n-1)}{1\times 2}{y}_n(-2)$$ $$-[x{y}_{n+1}+n{y}_n]=0$$ $$(1-x^2)y_{n+2}-2xn{y}_{n+1}-(n^2-n)y_n$$ $$-x{y}^{n+1}-n{y}_n$$ $$(1-x^2)y_{n+2}-2xn{y}_{n+1}-n^2{y}_n+\cancel{n{y}_n}$$ $$-x{y}_{n+1}-\cancel{n{y}_n}=0$$ $$(1-x^2)y_{n+2}-(2n+1)x{y}_{n+1}-n^2{y}_n=0$$ Hence Proved :)