Nth Derivative Of `e^{2x} Sin3x.Sin2x` | Leibnitz Theorem Of Nth Derivatives Examples

To Find n_th derivative of e^(2x) Sin3x Sin2x 

In this quick tutorial we will learn how to find n_th derivative of a function which is in the form of e^ax. sinbx. sincx. So let us understand steps to reach n_th derivative very carefully.

Solution: let             u = `e^(2x)`                     and                     v = `Sin3x .Sin2x`

Since we know that: 

`SinA.SinB` = `\[\frac{1}{2}\]` {cos(A-B) - cos(A+B)}

Now let:                 A = 3X            And                 B = 2X

Therefore,         `(A-B)` = `X`   and    `(A+B)` = `5X`

Also      `y = e^{x}.\frac{1}{2}.{cos(A-B) - cos(A+B)}`

Or                  `y = \frac{1}{2}.{e^{x}.cosX - e^{x}.cos5X}`

              

`N_th` derivative of   `e^{Ax}.Cos(Bx+c)` is:  

                   =  `(A^{2}+B^{2})^{n/2}.e^{Ax}.Cos(Bx+C+n.tan^{-1}\frac{B}{A})`

So `n_th`  derivative of   `\frac{1}{2}.[e^{2x}.CosX\]` is:

= `\frac{1}{2}.(2^{2}+1^{2})^{n/2}. e^{2x}. Cos(X+n.tan^{-1}\frac{1}{2})`

=   `\frac{1}{2}.5^{2}. e^{2x}.Cos(X+n.tan^{-1}\frac{1}{2})`

 

Similarly `n_th` derivative of `e^{2x}.Cos5X` as:

=   `\frac{1}{2}.29^{n/2}. e^{2x}.Cos(5X+n.tan^{-1}\frac{5}{2})`

Finally `N_th` derivative of  `e^(2x) Sin3x Sin2x` is:

=  `\frac{1}{2}.5^{2}. e^{2x}.Cos(X+n.tan^{-1}\frac{1}{2})`  -   `\frac{1}{2}.29^{n/2}. e^{2x}.Cos(5X+n.tan^{-1}\frac{5}{2})`