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Derivative of Inverse Trigonometric Functions

Derivative of the Arcsine

1



cos y would be adequate for the derivative of ^xsin



^{y , but we}

require the derivative of ^ysin1



^x . Therefore, our answer must be in terms of x.

By applying similar techniques, we obtain the rules for derivatives of Inverse Trigonometric Functions.

Rules:











1

2

1

2

1

2

1

2

1

2

sin 1 , 1

1

cos 1 , 1

1

tan 1 ,

1

cot 1 ,

1

sec 1 , 1

1

d du

u u

dx u dx

d du

u u

dx u dx

d du

dx u u dx

d du

dx u u dx

d du

u u

dx u u dx











 



  



 

 



 



 

 







1

2

2

sin sin

sin 1 cos

1 cos

1 1 sin

1 1

y x

x y

d d

x y

dx dx

y dy dx dy

dx y

y

x

 



  





  

 

Example

Find ^dy

dx if ^ysin1

 

^x3

Solution:

 

2 2

2 6

3

1 3

3 1 1

dy x

dx x x x

  

 

Example

A particle moves along the x-axis so that its position at any time ^t0 is



^{tan 1}

 

x t  ^ t . What is the velocity of the particle when ^t16?

Solution:

  

1

 

2

tan 1

1

1 1

1 2

d d

v t t t

dt t dt

t t

    

      

 



When t=16, the velocity is:

 

^{16 1 1 1}

1 16 2 16 136

v   



Example

Determine the derivative of



^sec1



^x



²

Solution:

    ^   ^ 

  

1 2 1 1

1

2

sec 2 sec sec

2 sec 1

1

d d

x x x

dx dx

x

x x

  





 

Example

Find the derivative of ^{f x}



^{arctan 2}



^x2x



Solution:

    

 

2 2 2

2 2

1 2

1 2

4 1

1 2

f x d x x

x x dx x

x x

  

 

 

 

Example

Determine all points where the tangent line to sin ^{1 1} ₂ y 2

x

 

  is horizontal.

Solution:

 

²

2 2

2

1 2

1 2

1 2

dy x

dx x

x

  

  

 

Now the derivative equals zero (line is horizontal) only when x=0.

We now need only the y-value.

Therefore



1

2

1

sin 1

2 0

sin 1 2 6 y







 

  

   

 



The point

 

^{, 0,}

x y _{ }6

  