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Thought-Process to Discover Knowledge

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User Guide

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exercise provides practice problems to become fluent in the concepts.

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summary of this topic

### Standard Results in Derivatives

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»  derivatives: standard results for algebraic expressions
d/(dx) a = 0

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

### Derivatives of Algebraic Expressions

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.

Keep tapping on the content to continue learning.
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"

Progress

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"

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