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mathsDifferential CalculusStandard Results in Derivatives

Derivatives of Inverse Trigonometric Functions

In this page, the derivatives of inverse trigonometric functions such as `arcsin`, `arccos`, etc. are discussed.



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Finding the derivative of `y = arcsinx`:
`y= arcsin x`
`sin y = x`
differentiate this
`cos y (dy)/(dx) = 1`

`(dy)/(dx) = 1/(cos y)`
` = 1/sqrt(1-sin^2 y)`
` = 1/sqrt(1-x^2)`

What does the above prove?

  • `d/(dx)arcsinx = 1/(sqrt(1-x^2))`
  • `d/(dx)arcsinx = 1/(sqrt(1-x^2))`
  • `d/(dx)sin x = 1/(sqrt(1-x^2))`

The answer is "`d/(dx)arcsinx = 1/(sqrt(1-x^2))`"

Finding the derivative of `y = arccosx`:
`y=arccosx`
`cos y = x`
differentiate this
`-sin y (dy)/(dx) = 1`

`(dy)/(dx) = 1/(-sin y)`
`(dy)/(dx) = (-1)/sqrt(1-cos^2 y)`
`(dy)/(dx) = (-1)/sqrt(1-x^2)`

What does the above prove?

  • `d/(dx)cosx = (-1)/(sqrt(1-x^2))`
  • `d/(dx)arccosx = (-1)/(sqrt(1-x^2))`
  • `d/(dx)arccosx = (-1)/(sqrt(1-x^2))`

The answer is "`d/(dx)arccosx = (-1)/(sqrt(1-x^2))`"

Finding the derivative of `y = arctanx`:
`y=arctanx`
`tan y = x`
differentiate this
`sec^2y (dy)/(dx) = 1`

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

What does the above prove?

  • `d/(dx)tanx = 1/(1+x^2)`
  • `d/(dx)arctanx = 1/(1+x^2)`
  • `d/(dx)arctanx = 1/(1+x^2)`

The answer is "`d/(dx)arctanx = 1/(1+x^2)`"

Finding the derivative of `y = arc secx`:
`y=arc secx`
`y` is in first or second quadrant.
`sec y = x`
differentiate this
`sec y tan y (dy)/(dx) = 1`
`(dy)/(dx) = 1/(sec y tan y)`
`sec y tan y` is always positive for values in first and second quadrant.
`(dy)/(dx) = 1/(sec y sqrt(sec^2 y-1))`
` = 1/(|x|sqrt(x^2-1))`

What does the above prove?

  • `d/(dx)arc sec x = 1/(|x|sqrt(x^2-1))`
  • `d/(dx)arc sec x = 1/(|x|sqrt(x^2-1))`
  • `d/(dx) sec x = 1/(|x|sqrt(x^2-1))`

The answer is "`d/(dx)arc secx = 1/(|x|sqrt(x^2-1))`"

Finding the derivative of `y = arc cscx`:
`y=arc cscx`
`y` is in range `-pi/2<=y<=pi/2`.
`csc y = x`
differentiate this
`-csc y cot y (dy)/(dx) = 1`

`(dy)/(dx) = (-1)/(csc y cot y)`
`csc y cot y` is always positive for values in range `-pi/2<=y<=pi/2`
`= (-1)/(csc y sqrt(csc^2 y-1))`
`= (-1)/(|x|sqrt(x^2-1))`

What does the above prove?

  • `d/(dx)arc cscx = (-1)/(|x|sqrt(x^2-1))`
  • `d/(dx)arc cscx = (-1)/(|x|sqrt(x^2-1))`
  • `d/(dx) cscx = (-1)/(|x|sqrt(x^2-1))`

The answer is "`d/(dx)arc cscx = (-1)/(|x|sqrt(x^2-1))`"

Finding the derivative of `y = arc cotx`:
`y=arc cotx`
`cot y = x`
differentiate this
`-csc^2y (dy)/(dx) = 1`

`(dy)/(dx) = (-1)/(csc^2y)`
`= (-1)/sqrt(1+tan^2 y)`
`= (-1)/sqrt(1+x^2)`

What does the above prove?

  • `d/(dx)arc cotx = (-1)/(1+x^2)`
  • `d/(dx)arc cotx = (-1)/(1+x^2)`
  • `d/(dx)cotx = (-1)/(1+x^2)`

The answer is "`d/(dx) arc cotx = (-1)/(1+x^2)`"

Derivatives of inverse trigonometric functions :
`d/(dx) arcsinx = 1/(sqrt(1-x^2))`
`d/(dx) arccosx = (-1)/(sqrt(1-x^2))`
`d/(dx) arctanx = 1/(1+x^2)`
`d/(dx) arc secx = 1/(|x|sqrt(x^2-1))`
`d/(dx) arc cscx = (-1)/(|x|sqrt(x^2-1))`
`d/(dx) arc cotx = (-1)/(1+x^2)`

Derivatives of inverse trigonometric functions : `text(trig fn) (y)=x`, so derivative in terms of x

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

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

`d/(dx) arctanx = 1/(1+x^2)`

`d/(dx) arc secx = 1/(|x|sqrt(x^2-1))`

`d/(dx) arc cscx = (-1)/(|x|sqrt(x^2-1))`

`d/(dx) arc cotx = (-1)/(1+x^2)`

Solved Exercise Problem:

Find the derivative of `(1+x^2)arctan x`

  • `1`
  • `1+2x arctan x`
  • `1+2x arctan x`

The answer is "`1+2x arctan x`"

`[ (1+x^2)arctan x]′`
`(1+x^2) (arctan x)′ + (1+x^2)′ arctan x`
`(1+x^2) (1/(1+x^2)) + 2x arctan x`
`1+2x arctan x`

                            
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