Introduction:

The integral calculus is concerned with the inverse problem of the given derivative of a function to find the function. In symbol, we require to find f(x) where, `d/dx` f(x) = g(x).In the trigonometric calculus the derivative of a constant is zero. There is no exact value for the integral. The derivative of a constant is zero. Calculus is widely used in mathematics, science, and engineering .it is useful for many fields and many other mathematical areas.

 

Derivative formulas:

 

1. `d / dx` (x n ) = n xn-1

2.   ` d/dx ` (sin x) = cos x

3.   ` d/dx` (cos x) = − sin x

4.  `d/dx ` (tan x) = sec2x

5.  ` d/dx` (cosec x) = −cosec x.sec x

6.   ` d/dx` (sec x) = sec x.tan x

7.   ` d/dx` (cot x) = −cosec2 x

                    Here, Inverse of sin can be written as sin-1x  = `1/sin x`

        We know the inverse trigonometric relation   `1/sin x` = cosec x  

 

Inverse derivative of sin x

 

Find  the derivative :  `d/dx ` `(sin^(-1)x)`  

  Solution:

             Let y = sin-1 x, then     sin y = x.

         Differentiate with respect to x . So we get  cos y  `(dy)/(dx)` =1

         Divided by cos y on both side So, we get       ` cos y / cos y` `(dy)/(dx)` = ` 1/cos y`

                                   `(dy)/(dx)` = ` 1/cos y``1/ sqrt(1-sin^2y)`                                        we know sin2x + cos2x = 1   So, cos x = `sqrt(1-sin^2x)`

                                             =` 1/sqrt(1-x^2)`                                                             we know sin y = x So, sin2y = x2

       So,   The inverse derivative of sin x i.e.,  `d/dx ` `(sin^(- 1)x)` = ` 1/sqrt(1-x^2)`

 

Other inverse trigonometric derivatives

           `d/dx ` `(cos^(- 1)x)``- 1/sqrt(1-x^2)`

               `d/dx ` `(tan^(- 1)x)`` 1/(1+ x^2)`

 

Examples of inverse derivatives:

 

Differentiate ` sqrt (sin x)` with respect to x:

 Solution:

         Let      y = ` sqrt (sin x)`

    u = x, v = sin u      So,that      y = v.

 `(du)/dx`  = 1 ,       `(dv)/(du)` = cos u,       `(dy)/(dv)` = (1/2) v-1/2

 Now            `(dy) / dx`  = [  `(dy)/(dv)`  ×   `(dv)/(du)` × `(du)/dx`]

                               = [(`(1/2)` v-1/2  × cos u × 1)]

                              = `[ (cos u ) / (2sqrt(v ))]`

                              = `[(cos x) / (2sqrt(sin x))] `

Answer:            `d/dx` ` sqrt (sin x)` = `[(cos x) / (2(sqrt sin x))] `