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Thought-Process to Discover Knowledge

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

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mathsDifferential CalculusStandard Results in Derivatives

Derivatives of Trigonometric Functions

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



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

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

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`

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

Solved Exercise Problem:

Find the derivative of `tan x -x`

  • `tan^2 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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