nubtrek

Server Error

Server Not Reachable.

This may be due to your internet connection or the nubtrek server is offline.

Thought-Process to Discover Knowledge

Welcome to nubtrek.

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

User Guide   

Welcome to nubtrek.

The content is presented in small-focused learning units to enable you to
  think,
  figure-out, &
  learn.

Just keep tapping (or clicking) on the content to continue in the trail and learn. continue

User Guide   

To make best use of nubtrek, understand what is available.

nubtrek is designed to explain mathematics and science for young readers. Every topic consists of four sections.

  nub,

  trek,

  jogger,

  exercise.

continue

User Guide    

nub is the simple explanation of the concept.

This captures the small-core of concept in simple-plain English. The objective is to make the learner to think about. continue

User Guide    

trek is the step by step exploration of the concept.

Trekking is bit hard, requiring one to sweat and exert. The benefits of taking the steps are awesome. In the trek, concepts are explained with exploratory questions and your thinking process is honed step by step. continue

User Guide    

jogger provides the complete mathematical definition of the concepts.

This captures the essence of learning and helps one to review at a later point. The reference is available in pdf document too. This is designed to be viewed in a smart-phone screen. continue

User Guide    

exercise provides practice problems to become fluent in the concepts.

This part does not have much content as of now. Over time, when resources are available, this section will have curated and exam-prep focused questions to test your knowledge. continue

summary of this topic

Standard Results in Derivatives

Standard Results in Derivatives

Voice  

Voice  



Home




 »  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)`


Derivatives of Inverse Trigonometric Functions

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

trek

Support Nubtrek     
 

You are learning the free content, however do shake hands with a coffee to show appreciation.
To stop this message from appearing, please choose an option and make a payment.




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


Keep tapping on the content to continue learning.
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".

we are not perfect yet...

Help us improve