nubtrek

Server Error

Server Not Reachable.

This may be due to your internet connection or the nubtrek server is offline.

Thought-Process to Discover Knowledge

Welcome to nubtrek.

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

User Guide   

Welcome to nubtrek.

The content is presented in small-focused learning units to enable you to
  think,
  figure-out, &
  learn.

Just keep tapping (or clicking) on the content to continue in the trail and learn. continue

User Guide   

To make best use of nubtrek, understand what is available.

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

  nub,

  trek,

  jogger,

  exercise.

continue

User Guide    

nub is the simple explanation of the concept.

This captures the small-core of concept in simple-plain English. The objective is to make the learner to think about. continue

User Guide    

trek is the step by step exploration of the concept.

Trekking is bit hard, requiring one to sweat and exert. The benefits of taking the steps are awesome. In the trek, concepts are explained with exploratory questions and your thinking process is honed step by step. continue

User Guide    

jogger provides the complete mathematical definition of the concepts.

This captures the essence of learning and helps one to review at a later point. The reference is available in pdf document too. This is designed to be viewed in a smart-phone screen. continue

User Guide    

exercise provides practice problems to become fluent in the concepts.

This part does not have much content as of now. Over time, when resources are available, this section will have curated and exam-prep focused questions to test your knowledge. continue

summary of this topic

Algebra of Differentiation

Algebra of Differentiation

Voice  

Voice  



Home



 »  Differentiation under Basic Arithmetic Operations

    →  `(a u)′ = a u′`

    →  `(u+v)′ = u′ + v′`

    →  `(u-v)′ = u′ - v′`

    →  `(uv)′=u′v+u v′`

    →  `(u/v)′=(u′v-u v′)/(v^2)`

Differentiation Under Basic Arithmetic

plain and simple summary

nub

plain and simple summary

nub

dummy

Derivative of a multiple of a function is multiple of the derivative of the function.

Derivative of a sum or difference is sum or difference of derivatives.

Derivative of product of two function is sum of derivatives of the functions scaled by the value of other function

simple steps to build the foundation

trek

simple steps to build the foundation

trek

Support Nubtrek     
 

You are learning the free content, however do shake hands with a coffee to show appreciation.
To stop this message from appearing, please choose an option and make a payment.




In this page, the derivatives of functions that are given as arithmetic operation of multiple functions is discussed. The arithmetic operations are addition, subtraction, multiplication, and division.


Keep tapping on the content to continue learning.
Starting on "Differentiation Under Basic Arithmetic". In this page, the derivatives of functions that are given as arithmetic operation of multiple functions is discussed. The arithmetic operations are addition, subtraction, multiplication, and division.

Derivative of a scalar multiple of function. Given `v(x)=color(deepskyblue)au(x)`.

`lim_(delta->0) (v(x+delta)-v(x))/delta`

`quad quad = lim_(delta->0) (color(deepskyblue)au(x+delta)-color(deepskyblue)au(x))/delta`

with continuity and differentiability conditions on `u` and `a`
`quad quad = color(deepskyblue)a lim_(delta->0) (u(x+delta)-u(x))/delta`

`quad quad = color(deepskyblue)a d/(dx) u`

what does the above prove?

  • `(a u)′ = a u′`
  • Derivative of a multiple of a function is multiple of the derivative of the function
  • Both the above

The answer is "both the above"

Intuitive understanding for
`(au)′ = a u′`
rate of change multiplies when the function is multiplied by a constant.

Finding derivative of sum or difference.

`lim_(delta->0) [(u(x+delta)+-v(x+delta))``-(u(x)+-v(x))]//delta`


`quad = lim_(delta->0) [(u(x+delta)-u(x)) ``+- (v(x+delta)-v(x))]//delta`


with continuity and differentiability conditions on `u` and `v`
`quad = lim_(delta->0) (u(x+delta)-u(x))/delta`

`+- lim_(delta->0) (v(x+delta)-v(x))/delta`


`quad quad = d/(dx) u +- d/(dx) v `

What does the above prove? `

  • `(u+v)′ = u′ + v′`
  • Derivative of a sum or difference is the sum or difference of derivatives.
  • `(u-v)′ = u′ - v′`
  • all the above

The answer is "all the above".

Intuitive understanding for
`(u+-v)′ = u′ +- v′`
rate of change adds (subtracts) when the function is added(subtracted).

Finding derivative of product.

`lim_(delta->0) [(u(x+delta) xx v(x+delta))``-(u(x) xx v(x))]//delta`


adding `color(coral)(u(x) xx v(x+delta))``-color(deepskyblue)(u(x) xx v(x+delta))`


`quad quad = lim_(delta->0) [u(x+delta) xx v(x+delta)``- color(deepskyblue)(u(x) xx v(x+delta))``+color(coral)(u(x) xx v(x+delta))``-u(x) xx v(x)]//delta`


with continuity and differentiability conditions on `u` and `v`
`quad quad = lim_(delta->0) [ color(deepskyblue)(v(x+delta)) xx (u(x+delta)``-color(deepskyblue)(u(x)))]//delta`


`quad + lim_(delta->0) [(color(coral)(u(x)) xx (color(coral)(v(x+delta))``-v(x)))]//delta`

`quad quad = v d/(dx) u + u d/(dx) v`

  • `(uv)′=u′v+u v′`
  • `(uv)′=u′v′`

The answer is "`(uv)′=u′v+u v′`"

Intuitive understanding of
`(uv)′=u′v+u v′`

Two functions are multiplied at every point. Rate of change of the product equals sum of

fixing one function `v` as a constant and rate of change of the other function `u′`

plus fixing the other function `u` as a constant and rate of change of the function `v′`

This is compared to `(au)′ = a u′`.

Finding derivative of division.
`lim_(delta->0) [``u(x+delta) -: v(x+delta)``-(u(x) -: v(x))]//delta`

`= lim_(delta->0) {`` 1/(v(x+delta)v(x))``xx [u(x+delta)v(x)`` - u(x)v(x+delta)]//delta }`


adding `color(coral)(u(x) v(x))-color(deepskyblue)(u(x) v(x))`

`= lim_(delta->0) {``1/(v(x+delta)v(x))`` [u(x+delta)v(x) -color(deepskyblue)(u(x) v(x))``- u(x)v(x+delta)+color(coral)(u(x) v(x))]//delta }`


`= lim_(delta->0) {``1/(v(x+delta)v(x))`` [v(x) (u(x+delta) -color(deepskyblue)(u(x)))``- u(x) (v(x+delta)- color(coral)(v(x)))]//delta }`


`= lim_(delta->0) {``1/(v(x+delta)v(x)) xx `` [v(x) (u(x+delta) -color(deepskyblue)(u(x)))]//delta)``- [u(x) (v(x+delta)- color(coral)(v(x)))]//delta }`

with continuity and differentiability conditions on `u` and `v`
`1/(v^2) ``xx {(du)/(dx) v`` - u (dv)/(dx)}`

What does the above prove?

  • `(u/v)′=(u′v-u v′)/(v^2)`
  • `(u/v)′=(u′)/(v′)`

The answer is "`(u/v)′=(u′v-u v′)/(v^2)`"

Intuitive understanding of `(u/v)′=(u′v-u v′)/(v^2)`

This can be rewritten as
`(u/v)′=``(u′ xx (1/v)`` + u xx (-1/v^2) xx v′`

The first part of the result is :
fixing `1/v` as a constant, rate of change of `u` is taken

The second part of the result is :
fixing `u` as a constant, rate of change of `1/v` is taken.
Rate of change of `1/v` is negative of rate of change of `v` and divided by `v^2`.

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

Derivative of a Multiple:
`(au)′ = a u′`

Derivative of Sum or Difference:
`(u+v)′ = u′ + v′`

`(u-v)′ = u′ - v′`

Derivative of Product: `(uv)′=u′v+u v′`

Derivative of Quotient: `(u/v)′=(u′v-u v′)/v^2`



           

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

Given `(dy)/(dx) = 2x^2` and `v=y/5`, what is `(dv)/(dx)`?

  • `4x`
  • `2/5 x^2`

The answer is "`2/5 x^2`"

`(dv)/(dx)`
`= d/(dx) y/5`
`= 1/5 d/(dx) y`
`=1/5 xx 2x^2`
`=2/5 x^2`

Given `(dy)/(dx) = sinx` and `v=y+20`, what is `(dv)/(dx)`?

  • `sin x + 20`
  • `sin x`

The answer is "`sin x`"

`(dv)/(dx)`
`=(d(y+20))/dx`
`=(dy)/(dx) + (d(20))/(dx)`
`=sin x + 0`

Given `y=x^2 sin x` what is `(dy)/(dx)`?
Note: `d/(dx) x^2 = 2x`
`d/(dx) sin x = cos x`

  • `2xcos x`
  • `2xsinx+x^2cosx`

The answer is "`2xsinx+x^2cosx`"

`(dy)/(dx)`
`=(d)/(dx) (x^2 sin x)`
applying product law of derivative `=sin x (d)/(dx)x^2 + x^2 (d)/(dx) sin x`
`=2xsinx+x^2cosx`

Given `y= (sin x)/(x^2)` what is `(dy)/(dx)`? Note: `d/(dx) x^2 = 2x`
`d/(dx) sin x = cos x`

  • `-2(cos x)/(x^3)`
  • `(x^2cosx-2xsinx)/(x^4)`

The answer is "`(x^2cosx-2xsinx)/(x^4)`"

`(dy)/(dx)`
`=(d)/(dx) [(sin x)/(x^2)]`
applying quotient law of derivative `=(x^2 d/(dx) sin x - sin x d/(dx)x^2 )/ (x^4)`
`=(x^2cosx-2xsinx)/(x^4)`

your progress details

Progress

About you

Progress

Derivative of a scalar multiple of function is given by first principles. What does the above prove.
u;you;whole;prime;equal
a times u, whole prime = a, u prime
derivative;multiple;function
Derivative of a multiple of a function is multiple of the derivative of the function
both;above
Both the above
The answer is "both the above"
Intuitive understanding for the given result is : rate of change multiplies when the function is multiplied by a constant.
Given d y by d x = 2 x squared and v = y by 5, what is d v by d x.
4
4 x
2;5;squared
2 by 5 x squared
The answer is "2 by 5 x squared"
Derivative of a multiple of a function is multiple of the derivative of the function.
Derivative of a Multiple: a u whole prime equals a u prime
Finding derivative of sum or difference is given by first principles. What does the above prove.
plus;equals;+;plus
derivative of; (u plus v) equals derivative of u ;plus; derivative of v
sum;difference
Derivative of a sum or difference is the sum or difference of derivatives.
minus;equals;minus;-
derivative of; (u minus v) equals derivative of u ;minus; derivative of v
all;above
all the above
The answer is "all the above".
Intuitive understanding for u plus or minus v whole prime equals u prime plus or minus v prime. Rate of change adds or subtracts when the function is added or subtracted.
Given d y by d x = sine x and v = y + 20, what is d v by dx.
1
2
The answer is "sine x"
Derivative of a sum or difference is sum or difference of derivatives.
Derivative of Sum or Difference: u plus or minus v whole prime equals u prime plus or minus v prime
Finding derivative of product of two functions is given by first principles. What does the above prove.
1
2
The answer is "u v whole prime = u prime v + u v prime"
Intuitive understanding of u v whole prime = u prime v + u v prime. Two functions are multiplied at every point. Rate of change of the product equals sum of ; fixing one function v as a constant and rate of change of the other function u prime; plus; fixing the other function u as a constant and rate of change of the function v prime. ;; This is compared with a u whole prime = a u prime.
Given y = x squared sine x, what is d y by d x.
1
2
The answer is "2 x sine x + x squared cos x"
Derivative of product of two function is sum of derivatives of the functions scaled by the value of other function
Derivative of Product: u v whole prime = u prime v + u v prime.
Finding derivative of division is given by first principles. What does the above prove.
1
2
The answer is " u divided by v whole prime = u prime v minus u v prime divided by v squared"
Intuitive understanding of, u divided by v whole prime = u prime v minus u v prime divided by v squared : This can be rewritten as u divided by v whole prime = u prime multiplied 1 by v, plus, u multiplied, minus 1 by v squared, multiplied v prime. ;; The first part of the result is, fixing 1 by v as constant, rate of change of u is taken. The second part of the result is, fixing u as a constant, rate of change of 1 by v is taken. Rate of change of 1 by v is negative of rate change of v divided by v squared.
Given y = sine x by x squared, what is d y by d x.
1
2
The answer is "x squared cos x minus 2 x sine x whole divided by x power 4"
Derivative of Quotient: u divided by v whole prime = u prime v minus u v prime divided by v squared

we are not perfect yet...

Help us improve