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Standard Results in Derivatives

Standard Results in Derivatives

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 »  Standard Results from derivatives of trigonometric functions
look for the pattern in the results and quickly work out them. No need to learn by rote
    `d/(dx) sin x = cos x`

    `d/(dx) cos x = -sin x`
`cos x = sin(pi/2 -x)` and the negative sign of x is carried in the result

    `d/(dx) tan x = sec^2 x`
`tan = (sin)/(cos)` and results in `1/(cos^2)`

    `d/(dx) cot x = -csc^2 x`
`cot = tan(pi/2-x)` and the negative sign of x is carried

    `d/(dx) sec x = sec x tan x`
`sec = 1/cos` and results in `sin/(cos^2)`

    `d/(dx) csc x = -csc x cot x`
`csc = sec(pi/2-x)` and the negative sign of x is carried

Derivatives of Trigonometric Functions

plain and simple summary

nub

plain and simple summary

nub

dummy

Derivatives of trigonometric functions
`d/(dx) sin x = cos x`
`d/(dx) cos x = -sin x`
`d/(dx) tan x = sec^2 x`
`d/(dx) cot x = -csc^2 x`
`d/(dx) sec x = sec x tan x`

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trek

simple steps to build the foundation

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In this page, the derivatives of trigonometric functions sin, cos, tan etc. are discussed.


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Starting on learning " Standard Results in Derivatives of Trigonometric Functions". In this page, the derivatives of trigonometric functions sine, cosine, tan et cetera are discussed.

Finding the derivative of `y=sin x` in first principles:

`(dy)/(dx)`

`=lim_(delta->0) [color(coral)(f(x+delta))``- color(deepskyblue)(f(x))]//delta`

`=lim_(delta->0) [color(coral)(sin(x+delta)``- color(deepskyblue)(sinx)]//delta`

`=lim_(delta->0) [color(coral)(sinx cos delta + cos x sin delta)``-color(deepskyblue)(sinx)]//delta`

`= lim_(delta->0) [color(coral)(cos x sin delta)]/delta`` - lim_(delta->0) (color(deepskyblue)(sin x) - color(coral)(sin x cos delta))/delta`

`= cos x lim_(delta->0) (sin delta)/delta`` - sin x lim_(delta->0)(1 - cos delta)/delta`

applying the standard limits
`= cos x xx 1 - sin x xx 0`

What does the above prove?

  • `d/(dx) sin x = cos x`
  • `d/(dx) sin x = cos x + sin x`

The answer is "`d/(dx) sin x = cos x`"

Finding the derivative of `y=cos x`:

`(d)/(dx) cos x`

`=d/(dx) sin(pi/2-x)`

applying chain rule of differentiation with `u=pi/2-x`
`=d/(du)sin(u) d/(dx) (pi/2-x)`

`=cos(u) xx (-1)`

`=-cos(pi/2-x)`

`=-sin(x)`

What does the above prove?

  • `d/(dx) cos x = sin x`
  • `d/(dx) cos x = -sin x`

The answer is "`d/(dx) cos x = -sin x`"

Finding the derivative of `y=tan x`

`(tan x)′`

`=((sin x)/(cos x))′`

applying quotient rule of differentiation
`=(cos x(sin x)′ - sin x(cos x)′)/(cos^2 x)`

`=(cos x(cos x) - sin x(-sin x))/(cos^2 x)`

`=(cos^2 x + sin^2 x)/(cos^2 x)`

`=1/(cos^2 x)`

`=sec^2(x)`

What does the above prove?

  • `d/(dx) tan x = sec^2 x`
  • `d/(dx) tan x = cos^2 x`

The answer is "`d/(dx) tan x = sec^2 x`"

Finding the derivative of `y=cot x`

`(cot x)′`

`=((cos x)/(sin x))′`

applying quotient rule of differentiation
`=(sin x(cos x)′ - cos x(sin x)′)/(sin^2 x)`

`=(sin x(-sin x) - cos x(cos x))/(sin^2 x)`

`=(-sin^2 x - cos^2 x)/(sin^2 x)`

`=(-1)/(sin^2 x)`

`=-csc^2(x)`

What does the above prove?

  • `d/(dx) cot x = cos^2 x`
  • `d/(dx) cot x = -csc^2 x`

The answer is "`d/(dx) cot x = -csc^2 x`"

Finding the derivative of `y=sec x`

`(sec x)′`

`=(1/(cos x))′`

applying quotient rule of differentiation
`=(cos x(1)′ - (cos x)′)/(cos^2 x)`

`=(cos x(0) - (-sin x))/(cos^2 x)`

`=(sin x)/(cos^2 x)`

`=sec x tan x`

What does the above prove?

  • `d/(dx) sec x = sec x`
  • `d/(dx) sec x = sec x tan x`

The answer is "`d/(dx) sec x = sec x tan x`"

Finding the derivative of `y=csc x`

`(csc x)′`

`=(1/(sin x))′`

applying quotient rule of differentiation
`=(sin x(1)′ - (sin x)′)/(sin^2 x)`

`=(sin x(0) - (cos x))/(sin^2 x)`

`=(-cos x)/(sin^2 x)`

`=-csc x cot x`

What does the above prove?

  • `d/(dx) csc x = -csc x`
  • `d/(dx) csc x = -csc x cot x`

The answer is "`d/(dx) csc x =- csc x cot x`"

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

Derivatives of Trigonometric Functions:
  `d/(dx) sin x = cos x`

  `d/(dx) cos x = -sin x`
`cos x = sin(pi/2 -x)` and the negative sign of x is carried in the result

  `d/(dx) tan x = sec^2 x`
`tan = (sin)/(cos)` and results in `1/(cos^2)`

  `d/(dx) cot x = -csc^2 x`
`cot = tan(pi/2-x)` and the negative sign of x is carried

  `d/(dx) sec x = sec x tan x`
`sec = 1/cos` and results in `sin/(cos^2)`

  `d/(dx) csc x = -csc x cot x`
`csc = sec(pi/2-x)` and the negative sign of x is carried



           

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

Find the derivative of `tan x -x`

  • `tan^2 x`
  • `cot^2 x`

The answer is "`tan^2 x`".

`tan x -x`
`sec^2 x -1`
`tan^2x`

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Progress

Finding the derivative of y = sine x in first principles is given. What does the above prove.
1
2
The answer is "d by dx of sine x = cos x "
Finding derivative of y= cosine x is given. What does the above prove.
1
2
The answer is "d by dx of cos x = minus sine x "
Finding derivative of y= tan x is given. What does the above prove.
1
2
The answer is "d by dx of tan x = secant squared x "
Finding derivative of y= cot x is given. What does the above prove.
1
2
The answer is "d by dx of cot x = minus co-secant squared x "
Finding derivative of y= secant x is given. What does the above prove.
1
2
The answer is "d by dx of secant x = secant x tan x"
Finding derivative of y= co-secant x is given. What does the above prove.
1
2
The answer is "d by dx of co-secant x = minus co-secant x cot x"
Derivatives of trigonometric functions is given.
Derivatives of trigonometric functions is listed. Look for the pattern and derive each one. As one uses them more,
Find the derivative of tan x minus x
tan
tan squared x
cot
cot squared x
The answer is "tan squared x"

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