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Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

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» Standard Results from derivatives of inverse trigonometric functions

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

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

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

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

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

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

*plain and simple summary*

nub

*plain and simple summary*

nub

dummy

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)`

*simple steps to build the foundation*

trek

*simple steps to build the foundation*

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

Starting on learning "Standard Results : Derivatives of Inverse Trigonometric Functions". In this page, the derivatives of inverse trigonometric functions such as sine inverse, cosine inverse, et cetera are discussed.

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)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))`

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)`

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

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

*comprehensive information for quick review*

Jogger

*comprehensive information for quick review*

Jogger

dummy

**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)`

*practice questions to master the knowledge*

Exercise

*practice questions to master the knowledge*

Exercise

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

- `1`
- `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`

*your progress details*

Progress

*About you*

Progress

Finding the derivative of y = sine inverse x or arc sine x is given. What does the above prove.

1

2

The answer is "d by d x of sine inverse x = 1 by square root 1 minus x squared"

Finding the derivative of y = cos inverse x or arc cosine is given. What does the above prove.

1

2

The answer is "d by d x of cos inverse x = minus 1 by square root 1 minus x squared"

Finding the derivative of y = tan inverse x or arc tan x is given. What does the above prove.

1

2

The answer is "d by d x of tan inverse x = 1 by 1 + x squared"

Finding the derivative of y = secant inverse x is given. What does the above prove.

1

2

The answer is "d by d x of secant inverse x = 1 by mod x, square root x squared minus 1"

Finding the derivative of y = co-secant inverse x or arc co-secant is given. What does the above prove.

1

2

The answer is "d by dx of cosec inverse x = minus 1 by mod x, square root of x squared minus 1"

Finding the derivative of y = cot inverse x or arc cot x is given. What does the above prove.

1

2

The answer is "d by dx of cot inverse x = minus 1 by 1 + x squared"

Derivatives of inverse trigonometric functions are listed.

Derivatives of the inverse trigonometric functions are given. Remember that the inverse functions are converted to trigonometric function on the other side of equation. So there is pattern in each of the results.

Find the derivative of the given function.

1

2

The answer is "1 + 2 x tan inverse x".