Trigonometric derivative formulas

Introduction :

Calculus is a branch of mathematics. Differentiation is a method to compute the rate at which a dependent output y changes with respect to the change in the independent input x. This rate of change is called thederivative of y with respect to x. In more precise language, the dependence of y upon x means that y is a function of x. The derivative of y with respect to x can be denoted as, `(dy)/(dx).`

                                              Source Wikipedia.


Trigonometric derivative formulas:


  Some derivative formulas are helped to solve the trigonometric functions. there are,

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 . cot x

6.  `d/dx ` `(sin^(- 1)x)` = ` 1/sqrt(1-x^2)`

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

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

                    These formulas are helped to find the derivative of trigonometric function.


Derivative of trigonometric functions help problems:


Derivative of trigonometric functions help problem 1:

 Find the derivative of  trigonometric function  y = 10x2 + 5tan x

  Solution:

              Let   y = 10x2 + 5tan x

                      `(dy)/dx` = `d/dx` [10x2 + 5tan x]

                          = 20x `(dx) / dx ` + 5`d/dx`(tan x)               (we know `d/dx`(tan x) = sec2x)

                    = 20x + 5 sec2x  

  Answer:  The derivative of given trigonometric function is  20x+5 sec2x


Derivative of trigonometric functions help problem 2:

  Find the derivative of given trigonometric  function  x tan x  with respect to x .

  Solution:

            Given trigonometric function is  x  tan x

                              Let  y  = x  tan x

                                  and  u =  x  and       v = tan x

                                    `(du)/(dx)` = 1   and    `(dv)/(dx)` = `sec^2x`

  The derivative multiplication rule of trigonometric function

                                            `d/dx(uv)` = `u (dv)/dx` + `v (du)/dx`

                                                ` (dy)/(dx)` = x `sec^2x` + tan x  1

                                                =[tan x + x sec2x]

            `d/dx`(x tan x)   = [tan x + x sec2x]

   Answer:  `d/dx`(x tan x)  = [tan x + x sec2x]

Derivative of trigonometric functions help problem 3:

  Find the derivative of  given function  y = 8x3 + 2cos x

  Solution:

                          Let  y = 8x3 +  2cos x

                              `(dy)/dx` = `d/dx` [8x3 + 2cos x]

                                = (8 × 3)x2 `(dx) / dx ` + `d/dx`(2cos x)               (we know `d/dx`(cos x) = - sin x)

                            = 24x2 +  (-2sin x)

                                = 24x2 - 2sin x

  Answer: Derivative of given function 8x3 +  2cos x is 24x2 - 2sin x