Revision:Chain, Product and Quotient - The Student Room
The Student Room

Revision:Chain, Product and Quotient

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Chain, Product and Quotient

Contents

The Chain Rule

The chain rule is very important in differential calculus and states that:

\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}

This rule allows us to differentiate a vast range of functions.

Proof

Suppose that y is a function of t, and t is a function of x.

If \delta y, \delta t and \delta x are corresponding increments in the variables in the variables y, t andx, then

\frac{\delta y}{\delta x}=\frac{\delta y}{\delta t}\times \frac{\delta t}{\delta x} (1)

When \delta y, \delta t and \delta x tend to zero,

\frac{\delta y}{\delta x} \rightarrow \frac{dy}{dx}

\frac{\delta y}{\delta t} \rightarrow \frac{dy}{dt}

\frac{\delta t}{\delta x} \rightarrow \frac{dt}{dx}

so equation (1) becomes

\frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}

Examples

If y=(1+x^2)^3, find \frac{dy}{dx}.

Let t=1+x^2

\frac{dt}{dx}=2x

y=t^3

\frac{dy}{dt}=3t^2

By the Chain Rule, \frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}.

\Rightarrow \frac{dy}{dx}=3t^2 \times 2x= 3(1 + x^2)^2 \times 2x

\frac{dy}{dx}=6x(1 + x^2)^2

In examples such as the above one, with practice it should be possible for you to be able to simply write down the answer without having to let t=1 + x^2 etc. This is because:

When y=(f(x))^n

\frac{dy}{dx}=f'(x)\times n \times (f(x))^{(n-1)}

In other words, the differential of something in a bracket raised to the power of n is the differential of the bracket, multiplied by n multiplied by the contents of the bracket raised to the power of (n-1).


The Product Rule

This is another very useful formula, when we have two functions u and v, multiplied together:

\frac{d}{dx}(uv)=v\frac{du}{dx}+u\frac{dv}{dx}

Proof

Consider y=f(x) \times g(x).

Then y=uv (1)

Let x increase by a small amount \delta x, with corresponding increases in y, u and v of \delta y, \delta u and \delta v, so that

y+\delta y=(u+\delta u)(v+\delta v)

\Rightarrow y + \delta y=uv+u\delta v+v\delta u+\delta u\delta v (2)

Then, subtracting (1) from (2).

\delta y = u\delta v + v\delta u+\delta u\delta v

Then, dividing by \delta x,

\frac{\delta y}{\delta x}=\frac{u\delta v}{\delta x}+ \frac{v\delta u}{\delta x}+\frac{\delta u\delta v}{\delta x}

\Rightarrow \frac{\delta y}{\delta x}=u\frac{\delta v}{\delta x}+v\frac{\delta u}{\delta x}+\delta u\frac{\delta v}{\delta x}

Let \delta x \rightarrow 0.

Then \delta u \rightarrow 0 and \delta v \rightarrow 0,

\frac{\delta y}{\delta x} \rightarrow \frac{dy}{dx}

\frac{\delta v}{\delta x} \rightarrow \frac{dv}{dx}

\frac{\delta u}{\delta x} \rightarrow \frac{du}{dx}

\Rightarrow \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}+0\times\frac{dv}{dx}

\Rightarrow \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}

Examples

Differentiate x(x^2+1)

Let u=x and v=x^2+1

\Rightarrow \frac{du}{dx}=1 and \frac{dv}{dx}=2x

\frac{d}{dx}(x(x^2+1))=(x^2+1)\times 1 + x\times 2x

\frac{d}{dx}(x(x^2+1))=x^2+1+2x²

\Rightarrow \frac{d}{dx}(x(x^2+1))=3x^2+1

Again, with practice you shouldn't have to write out u = ... and v = ... every time.

The Quotient Rule

This formula lets us differentiate two functions divided by each other.

\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Proof

Writing u=f(x) and v=g(x), then

y=\frac{u}{v} (1)

and

y+\delta y=\frac{u+\delta u}{v+\delta v} (2)

Substituting (1) into (2),

\delta y=\frac{u+\delta u}{v+\delta v}-\frac{u}{v}

\delta y=\frac{v(u+\delta u)-u(v+\delta v)}{v(v+\delta v)}

\delta y=\frac{vu+v\delta u-uv-u\delta v}{v^2+v\delta v}

Dividing by \delta x,

\frac{\delta y}{\delta x}=\frac{v\frac{\delta u}{\delta x}-u\frac{\delta v}{\delta x}}{v^2+v\delta v}

Let \delta x \rightarrow 0, then \delta u \rightarrow 0, \delta v \rightarrow 0 and

\frac{\delta y}{\delta x}\rightarrow \frac{dy}{dx}

\frac{\delta v}{\delta x}\rightarrow \frac{dv}{dx}

\frac{\delta u}{\delta x}\rightarrow \frac{du}{dx}

\Rightarrow \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Examples

If y=\frac{x^3}{x+4}, find \frac{dy}{dx}.

Let u=x^3 and v=x+4.

\Rightarrow \frac{du}{dx}=3x^2 and \frac{dv}{dx}=1

\frac{d}{dx}(\frac{x^3}{x+4})=\frac{(x+4)\times (3x^2)-(x^3)\times (1)}{(x+4)^2}

\Rightarrow \frac{d}{dx}(\frac{x^3}{x+4})=\frac{2x^3+12x^2}{(x+4)^2}


Comments

Retrieved from "http://www.thestudentroom.co.uk/wiki/Revision:Chain,_Product_and_Quotient"
Related Content Toggle
Loading...
Discussions Toggle
Warwick Maths/MORSE/Stats applicants and questions 2012
started by: TheTallOne
forum: University of Warwick
replies: 331
last post: 1 Minute Ago
TALL GIRLS and CLOTHING WOES
started by: what's my name
forum: Clothes and Accessories
replies: 130
last post: 1 Minute Ago
Type of music - whats yours?
started by: ellie_mitchell7
forum: Music
replies: 101
last post: 1 Minute Ago
help do we qualify for help
started by: jmfc
forum: Advice on Everyday Issues
replies: 11
last post: 1 Minute Ago
Free Representation Unit Tests
started by: Silphe
forum: Legal
replies: 4
last post: 1 Minute Ago
Your Motivation Song?
started by: PaperBoy
forum: Music
replies: 12
last post: 1 Minute Ago
Medical school desired personality
started by: potatohead
forum: Medicine
replies: 17
last post: 1 Minute Ago
Life is short, I want to be happy, but I pick my A-levels next week. God!
started by: Afootballmug
forum: A-Levels, ASs, A2s, VCEs
replies: 74
last post: 1 Minute Ago
LSE postgrad applicants 2012-13
started by: FinLuv
forum: Postgraduate
replies: 1289
last post: 1 Minute Ago
Oxford Graduate Application 2012/13
started by: Poincare
forum: Postgraduate
replies: 1076
last post: 1 Minute Ago
Most annoying thing people do at University?
started by: The Phelps
forum: Student Life
replies: 223
last post: 1 Minute Ago
HELP! LOST PASSPORT!! Cost of replacement?
started by: weirdwakkidolli
forum: Advice on Everyday Issues
replies: 1
last post: 1 Minute Ago
Italy?
started by: sylvie92
forum: International Lounge
replies: 579
last post: 1 Minute Ago
RAF Brize Norton - Potential Officer/Aircrew day
started by: NDW2009
forum: Armed Forces
replies: 2
last post: 1 Minute Ago
Muslims try to ban negative reporting of Islam in Britain
started by: DZJX
forum: Society
replies: 76
last post: 1 Minute Ago
Calling all 2012 VetMed entry hopefuls, *VERY* Early i know!
started by: JLW
forum: Veterinary Medicine
replies: 6248
last post: 1 Minute Ago
Sherlock (BBC TV series)
started by: Ape Gone Insane
forum: TV
replies: 2866
last post: 1 Minute Ago
Second racist gang attacks on white man by "asians"
started by: Zeffy
forum: News and Current Affairs
replies: 33
last post: 1 Minute Ago
Didn't sit GCSEs now 18, completely stuck
started by: Troglet
forum: Apprenticeships and alternatives to University
replies: 10
last post: 1 Minute Ago
Lost 10lb's in 13 days
started by: ras90
forum: Fitness
replies: 9
last post: 2 Minutes Ago
Article Updates Toggle
Modern Language Offers 2012
edited by: Strüberry, 5 Minutes Ago
Law Offers 2012
edited by: Natmay93, 7 Minutes Ago
Medicine Applicants Stalking Page 2012 Entry
edited by: ctenses, 12 Minutes Ago
Geography Offers 2012
edited by: bm127, 53 Minutes Ago
Pharmacy Offers 2012
edited by: qasidb, 1 Hour Ago
LSE Applicants Stalking Page 2012 Entry
edited by: BigBlackDog, 2 Hours Ago
Neuroscience Offers 2012
edited by: Mattyphillips5320, 2 Hours Ago
Lancaster Applicants Stalking Page 2012 Entry
edited by: vedderfan94, 3 Hours Ago
KCL 2012 Applicants Stalking Page
edited by: chocopops, 3 Hours Ago
Oxford Applicants Stalking Page 2012 Entry
edited by: chocopops, 3 Hours Ago
Contact Us | Site Rules | Staying Safe on TSR | Advertising | Staff Blog | Essays & Coursework | Terms & Conditions | Top
Customise your TSR | Life Advice | Hobbies and Interests | Debate and Current Affairs | Study Help | University and University courses
Universities and HE Colleges | Careers, Employment and Gap Years | General Discussion

Customise your TSR