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

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

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

Tricky Logarithm Problem…

Find the exact solution to the equation ${ 3 }^{ x }{ e }^{ 7x+2 }=15$.

${ 3 }^{ x }{ e }^{ 7x+2 }=15\\ \\ { 3 }^{ x }{ e }^{ 7x }{ e }^{ 2 }=15\\ \\ { 3 }^{ x }{ e }^{ 7x }=\frac { 15 }{ { e }^{ 2 } } \\ \\ { \left( 3{ e }^{ 7 } \right) }^{ x }=\frac { 15 }{ { e }^{ 2 } } \\ \\ \log _{ e }{ \left( { \left( 3{ e }^{ 7 } \right) }^{ x } \right) } =\log _{ e }{ \left( \frac { 15 }{ { e }^{ 2 } } \right) } \\ \\ x\log _{ e }{ \left( 3{ e }^{ 7 } \right) } =\log _{ e }{ \left( \frac { 15 }{ { e }^{ 2 } } \right) } \\ \\ x=\frac { \log _{ e }{ \left( \frac { 15 }{ { e }^{ 2 } } \right) } }{ \log _{ e }{ \left( 3{ e }^{ 7 } \right) } } =\frac { \log _{ e }{ 15-\log _{ e }{ \left( { e }^{ 2 } \right) } } }{ \log _{ e }{ 3+\log _{ e }{ \left( { e }^{ 7 } \right) } } } =\frac { \ln { 15-2\log _{ e }{ \left( e \right) } } }{ \ln { 3+7\log _{ e }{ \left( e \right) } } } \\ \\ =\frac { \ln { 15-2 } }{ \ln { 3+7 } } ,\quad \therefore \quad x=\frac { -2+\ln { 15 } }{ 7+\ln { 3 } } \\$

Ordinary Graph Transformations:

$GRAPH\quad TRANSFORMATIONS:\\ \\ f\left( x+a \right) ,\quad -a\quad x\quad coordinates.\\ \\ f\left( x-a \right) ,\quad +a\quad x\quad coordinates.\\ \\ f(ax),\quad Multiply\quad x\quad coordinates\quad by\quad \frac { 1 }{ a } .\\ \\ f\left( x \right) +a,\quad +a\quad y\quad coordinates.\\ \\ f\left( x \right) -a,\quad -a\quad y\quad coordinates.\\ \\ af\left( x \right) ,\quad Multiply\quad y\quad coordinates\quad by\quad a.$

(a+b)(a-b)=a^2-b^2 (The Real Proof) (Difference Of Two Squares Demystified)

Ever wondered why (a+b)(a-b)=a^2-b^2? The videos below will reveal to you why we accept this fact. Enjoy!

1) Why (a+b)(a+b)=a^2+2ab+b^2

2) Why (a+b)(a-b)=a^2-b^2

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