Nth Derivative `Sin^(4)x` | Successive Differentiation | Leibnitz Theorem Solved Example

Nth derivative of Sin^(4)x

In this quick tutorial you will learn that how to find the nth derivative of a trigonometric function like Sin(x) or Sin^4(x) by successive differentiation method. So observe each and every step very carefully and meanwhile if you have any doubt related to the solution at any step please leave us a comment we assure you that you will hear from us as soon as possible. So let’s now…..

 

Given:                                  `y` =  `sin^(4)x`   

                             

                              Multiply and divide `sin^(4)x` by 4 we get

                                             `y` =  `\frac{(2sin^(2)x)^(2)}{4}`

Or                                         `y` =  `\frac{(1-cos2x)^(2)}{4}`            Since `(2sin^(2)x)^(2)=(1-cos2x)^(2)` 

Now                     expanding `(1-cos2x)^(2)` we get the following result

i.e                                      `y` =  `\[\frac{(1 + \cos ^2 2x - \2cos 2x)}{4}\]`

Again,                   multiply and divide the term `(cos2x)^(2)` with 4 we get the following result.

i.e.,                                      `y` =  `\[\frac{(1 + \frac{4\cos ^2 2x}{4} - \2.cos 2x)}{4}\]`

Or                                         `y` =  `\[\frac{1}{4} + \frac{1}{16} + \frac{1}{16}\(1+cos 4x)^2 - \frac{1}{2}\cos 2x\]`

Or                                         `y` =  `\[\frac{1}{4} + \frac{1}{16} + \frac{1}{16}\(1+cos^(2) 4x+2cos4x )- \frac{1}{2}\cos 2x\]`

Again Multiply & Divide `cos^(2)4x by 2 we get a result which is differentiable easily.

Or          `y` =  `\[\frac{1}{4} + \frac{1}{16} + \frac{1}{16}\(1+\frac{2cos^(2) 4x}{2}+2cos4x )- \frac{1}{2}\cos 2x\]`

Or          `y` =  `\[\frac{1}{4} + \frac{1}{16} + \frac{1}{16}\(1+\frac{1+cos8x}{2}+2cos4x )- \frac{1}{2}\cos 2x\]`

Or          `y` =  `\[\frac{1}{4} + \frac{1}{16} + \frac{1}{16}\1+\frac{1}{32}+\frac{cos8x}{32}+\frac{cos4x}{8} - \frac{1}{2}\cos 2x\]`

Now                     since `n_th` derivative of `sin(ax+b)` is:

                                             =  `[\a^(n). sin(ax+b+n.\frac{\pi }{2}]`

So,                        now writing `n_th` derivative of each term we get.

                                            = `\frac{1}{32}\8^(n). sin(8x+n.\frac{\pi }{2})`  - `\frac{1}{8}\4^(n). sin(4x+n.\frac{\pi }{2})` -  

`\frac{1}{2}{.2^n}.\cos (2x + n.\frac{\pi }{2})`