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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.

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

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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

Welcome to **nub****trek**.

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figure-out, &

learn.

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

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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 *

*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`

*simple steps to build the foundation*

trek

*simple steps to build the foundation*

trek

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

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`

*your progress details*

Progress

*About you*

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"