Derivative of Inverse Trigonometric Functions

 

Motivation:

                           Consider the equation  sin(x)= 0 ,  what would be the possible value of x ?  x can be 0, , 2 , 3 , or - ,…    The trigonometric functions are periodic so that we need to restrict the domain; it is so-called the principal value. . Thus, the inverse function for each trigonometric function can exist.   The inverse functions of trigonometric functions are defined as follows:

 

                      y= arcsin(x)    x=sin(y) ,    -   y  

                      y= arcos(x)     x=cos(y) ,  0  y  

                      y= arctan(x)    x=tan(y) ,  -  < y <

                      y= arccsc(x)    x=csc(y) ,      y   ,  y  0

                      y= arcsec(x)    x=sec(y),    0  y   .  y  

                      y= arccot(x)     x=cot(y),  -  < y < , y  0

 

                        When those things are clear, we will start to find the derivatives of those inverse functions.  Those functions are important because the results of some integrals will be represented by those functions in the following sections.

 

 

Derivative of arcsin(x)

                             Let          y = arcsin(x)  .

                             Thus, we have

                                            sin( arcsin(x) ) = x,  i.e.,   sin(y) = x

 

                             Differentiate the equality with respect to x on both sides:

 

                                      cos(y)  = 1         ( chain rule )

                            = 

 

                   So, the question turns out to be to find  cos(y) under the condition that x=sin(y) .  From one of trigonometric equalities,  the relationship between cosine and sine is

                           sin²(y) + cos²(y) = 1

 

                And at the beginning of this section,  y is confined in the range : -   y   In this range, cos(y) is always positive.  That implies

 

                                         cos(y)=  =

                                         =    

                 

                 In other words,

                                                      =  

 

 

                Similarly approach can be applied on other inverse trigonometric functions that we summarize as blow:

 

 

Theorem:

                (1)   =

                (2)   =

                (3)   =

                (4)    =

                (5)    =

                 (6)    =

 

          Proof:

(1)   is done just right before this theorem. We use it to prove the other two.

(2)    

We can use  arccos(x) =  - arcsin(x)

     = -

 

The other approach is to follow the similar way we do for arcsin(x):

                         y=arcos(x)

  cos(y) = x,  0  y  

 -sin(y)  = 1

So,

             =  =    ( Remember 0  y   )

 

                         (3)

                                   y=arctan(x)

  tan(y) = x

 sec²(x)  = 1

  = 

              We can use 

                          x= tan(y)= ,  sin²(y)+cos²(y) = 1

               cos²(y) =

So,

          =

 

                          (4)

                                          y= arcsrc(x)

                                    sec(y) = x

                                    sec(y)tan(y)  = 1

                                     =

 

                                                        We can use the following diagram :

                                                                 

                                           So,   sin(y)= ,  cos(y) =

                                     =

 

 

(5)      arccot(x) =  - arctan(x)

So,

              arccot(x) = -  arctan(x) =

 

(6)      arccsc(x) =  - arcsec(x)

So,

              arccsc(x) = -  arcsec(x) =