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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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» derivatives: standard results for algebraic expressions

`d/(dx) a = 0`

`d/(dx) x^n = n x^(n-1) `

*plain and simple summary*

nub

*plain and simple summary*

nub

dummy

`d/(dx) a = 0`

`d/(dx) x^n = n x^(n-1) `

*simple steps to build the foundation*

trek

*simple steps to build the foundation*

trek

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In this page, Derivatives of standard functions in algebraic expressions are explained.

Starting on learning "Standard Results : Derivatives of Algebraic Expressions". In this page, Derivatives of standard functions in algebraic expressions are explained.

Finding the derivative of `y=a` in first principles:

`(dy)/(dx)`

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

`=lim_(delta->0) [color(coral)(a)`` - color(deepskyblue)(a)]//delta`

`=lim_(delta->0) [0]//delta`

What does the above prove?

- `d/(dx) (a) = 0//0`: indeterminate value
- `d/(dx) (a) = 0`

The answer is "`d/(dx) (a) = 0`". Note that `delta` is very small value and the numerator is `0`. So the limit evaluates to `0`.

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

`(dy)/(dx)`

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

`=lim_(delta->0) [color(coral)((x+delta)^n)`` - color(deepskyblue)(x^n)]//delta` *using binomial expression for `(x+delta)^n`* `=lim_(delta->0) [color(coral)((x^n +ndelta x^(n-1)``+ ^nC_2 delta^2x^(n-2)``+ cdots+delta^n)`` - color(deepskyblue)(x^n)]//delta` * canceling `color(coral)(x^n)` and `color(deepskyblue)(-x^n)`*

`=lim_(delta->0) [color(coral)((ndelta x^(n-1)``+ ^nC_2 delta^2x^(n-2)``+ cdots+delta^n)``]//delta` * dividing all terms by `delta`*

`=lim_(delta->0) [color(coral)((nx^(n-1)``+ ^nC_2 delta^1x^(n-2)``+ cdots+delta^(n-1))``]` * applying limit `delta->0`*

What does the above prove?

- `d/(dx) (x^n) = nx^(n-1)`
- `d/(dx) (x^n) = nx^(n)`

The answer is "`d/(dx) (x^n) = nx^(n-1)`". The `delta` is substituted as `0` and only one term is non zero.

*comprehensive information for quick review*

Jogger

*comprehensive information for quick review*

Jogger

dummy

**Derivatives of Algebraic Expressions** :

Derivative of a constant

`d/(dx) a = 0`* rate of change of constant is 0*

Derivative of power:

`d/(dx) x^n = n x^(n-1) `*small change results in binomial expansion and only one factor remains*

*practice questions to master the knowledge*

Exercise

*practice questions to master the knowledge*

Exercise

Find the derivative of `x^3+4x^2-3x-2`

- `3x^2+8x-3-2`
- `3x^2+8x-3`
- `3x^3+8x^2-3x-2`

The answer is "`3x^2+8x-3`"

*your progress details*

Progress

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Progress

Finding derivative of y = a in first principles is given. What does the above prove.

1

2

The answer is "d by d x of a = 0". Note that delta is very small value and the numerator is 0 . So the limit evaluates to 0.

Finding derivative of y = x power n in first principles is given. What does the above prove.

1

2

The answer is "d by d x of x power n = n x power n minus 1"

Derivatives of algebraic expressions listed.

Derivatives of algebraic expressions: derivative of a constant is 0. Note that rate of change of constant is 0.;; derivative of power of a variable: d by d x of x power n = n x power n minus 1. Note that small change results in binomial expansion and only one factor remains.

Find the derivative of the given polynomial.

1

2

3

The answer is "3 x squared + 8 x minus 3"