# Proving The Quotient Rule…

$y=\frac { u }{ v } ,\quad \therefore \quad \left( y+\delta y \right) =\frac { \left( u+\delta u \right) }{ \left( v+\delta v \right) } \\ \\ \left( y+\delta y \right) \left( v+\delta v \right) =u+\delta u\\ \\ yv+y\delta v+v\delta y+\delta y\delta v=u+\delta u\\ \\ but...\quad yv=u,\\ \\ so...\\ \\ y\delta v+v\delta y+\delta y\delta v=\delta u\\ \\ \frac { u }{ v } \delta v+v\delta y+\delta y\delta v=\delta u\\ \\ u\delta v+{ v }^{ 2 }\delta y+v\delta y\delta v=v\delta u\\ \\ { v }^{ 2 }\delta y=v\delta u-u\delta v-v\delta y\delta v\\ \\ { v }^{ 2 }\frac { \delta y }{ \delta x } =v\frac { \delta u }{ \delta x } -u\frac { \delta v }{ \delta x } -v\delta v\frac { \delta y }{ \delta x } \\ \\ But\quad as\quad \delta x\rightarrow 0,\quad \frac { \delta y }{ \delta x } \rightarrow \frac { dy }{ dx } ,\quad \frac { \delta u }{ \delta x } \rightarrow \frac { du }{ dx } ,\quad \frac { \delta v }{ \delta x } \rightarrow \frac { dv }{ dx } \quad and\quad \delta v\rightarrow 0.\\ \\ So\quad you\quad get:\\ \\ { v }^{ 2 }\frac { dy }{ dx } =v\frac { du }{ dx } -u\frac { dv }{ dx } \\ \\ \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } }$

# Tricky Integration Problem

$\int _{ \frac { \pi }{ 3 } }^{ \frac { \pi }{ 2 } }{ 64\sin ^{ 2 }{ tcostdt } }$

$u=sint\quad \therefore \quad { u }^{ 2 }=\sin ^{ 2 }{ t } \\ \\ If\quad u=sint,\quad \frac { du }{ dt } =cost,\quad \therefore \quad du=costdt,\quad \therefore \quad \frac { 1 }{ cost } du=dt\\ \\ When\quad t=\frac { \pi }{ 2 } ,\quad u=sin\left( \frac { \pi }{ 2 } \right) =1.\quad When\quad t=\frac { \pi }{ 3 } ,\quad u=sin\left( \frac { \pi }{ 3 } \right) =\frac { \sqrt { 3 } }{ 2 } .\\ \\ So\quad we\quad must\quad integrate:\quad \int _{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }{ 64{ u }^{ 2 } } \cdot cost\cdot \frac { 1 }{ cost } du,\\ \\ which\quad is:\quad \int _{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }{ 64{ u }^{ 2 } } du\quad =\quad { \left[ \frac { 64{ u }^{ 3 } }{ 3 } \right] }_{ \frac { \sqrt { 3 } }{ 2 } }^{ 1 }\\ \\ =\left( \frac { 64{ \left( 1 \right) }^{ 3 } }{ 3 } \right) -\left( \frac { 64{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 3 } }{ 3 } \right) =\frac { 64 }{ 3 } -8\sqrt { 3 }$

# Trapezium Rule Formula – Derivation

Find out how to come up with the Trapezium Rule formula from scratch.

1) Derive the formula for the area of trapeziums:

2) Use the area of trapeziums formula to come up with the Trapezium Rule formula:

Related:

Derive the formula to find areas underneath curves

# FREE MATHS VIDEOS – FOR A LEVEL & GCSE STUDENTS

Welcome to MathsVideos.net. On this website you will find maths videos containing tricks and advice that you can use to pass your maths exams. MathsVideos.net was recently created in January 2014. These videos are produced by a student who is doing his A-Levels in the UK.

# Find The Formula For The Volume Of Cones

Find the volumes of cones formula using differentiation and integration. This playlist will show you how to come up with the volumes of cones formula from scratch. Knowledge about how to differentiate and integrate required.