# How to differentiate y=arctanx

Below I’m going to demonstrate how to integrate y=arctanx…

Firstly, we need to know that:

$\sin ^{ 2 }{ y } +\cos ^{ 2 }{ y } =1\\ \\ \frac { \sin ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } +\frac { \cos ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } =\frac { 1 }{ \cos ^{ 2 }{ y } } \\ \\ \tan ^{ 2 }{ y } +1=\sec ^{ 2 }{ y }$

We also need to know that:

$x=\tan { y } \\ \\ x=\frac { \sin { y } }{ \cos { y } } \\ \\ x\cdot \cos { y } =\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } +x\cdot \left( -\sin { y } \right) =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } -x\sin { y } =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +x\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +\frac { \sin ^{ 2 }{ y } }{ \cos { y } } \\ \\ \frac { 1 }{ \cos { y } } \cdot \frac { dx }{ dy } \cdot \cos { y } =\frac { 1 }{ \cos { y } } \left( \cos { y+\frac { \sin ^{ 2 }{ y } }{ \cos { y } } } \right) \\ \\ \frac { dx }{ dy } =1+\tan ^{ 2 }{ y } \\ \\ \therefore \quad \frac { dx }{ dy } =\sec ^{ 2 }{ y }$

And finally:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Now, using implicit differentiation:

$y=\arctan { x } \\ \\ \tan { y } =x\\ \\ \sec ^{ 2 }{ y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \sec ^{ 2 }{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \tan ^{ 2 }{ y+1 } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ { x }^{ 2 }+1 }$

This week’s updates are as follows:

You can learn how to draw a tesseract using the video below. This depiction of a tesseract can be found in the book ‘The Visual Guide To Extra Dimensions (Volume 1) – Visualizing the Fourth Dimension, Higher Dimensional Polytopes and Curved Hypersurfaces’ written by Chris McMullen Ph.D, a particle physicist based in the United States.

This next video contains a few perspective drawings and an animation of a sphere passing through a sheet of paper. It was made to demonstrate that perhaps the universe we inhabit is a construct of some kind.

You can explore non-euclidean space, or more specifically – a 3 dimensional sphere with lines passing through it both vertically and horizontally…

For those interested in seeing basic shapes rotate, you can watch a line, square and cube rotate about 360 degrees and additionally; a point, circle and sphere spin around an axis below…

Now, most of these videos were made in order to help me visualise a few concepts brought to light in Chris McMullen’s book. I hope you do get to enjoy watching these new clips. Feedback is appreciated as always. ðŸ™‚

# Maths Videos For Web News…

This month I managed to upload more proofs to my Google “A Level Maths Proofs” page which can be accessed through this link [https://plus.google.com/communities/106007058741903558109]

I also made a few new maths videos unrelated to proofs for those interested in mathematical art and funky ideas. I must admit, the quality of my videos (in terms of their resolution) weren’t that great, but hopefully you’ll like them.

My latest clip is called “Percentage Count (0% – 100%)“, designed for those who are struggling to understand what percentages really are. This video illustrates that one percent of something is really one part out of a hundred Â equal parts of an object. You require one hundred hundredths to produce 100% which is quite simply 1.00.

# Free A Level Mathematics Proofs

Today you can download and view my hand written A Level maths proofs. These hand written proofs can be found at:Â https://plus.google.com/communities/106007058741903558109

If you do decide to share my work, please make sure my web links remain intact so that I can continue to build and re-invest in mathsvideos.net.

As always, I strive to produce the clearest A Level maths proofs on the internet, so that you can pass your C1, C2, C3 and C4 exams with flying colours. I’m not the kind of tutor who simply tells students to plug numbers into formulas. I make sure my students understand the problems they’re dealing with – using proofs and logical models.

# Logarithmic Proof (23/03/2015)

Prove that:

$\ln { \left( xy \right) } =\ln { x+\ln { y } }$

Now, say that:

${ a }^{ m }=x\\ \\ \therefore \quad \log _{ a }{ x=m } \\ \\ { a }^{ q }=y\\ \\ \therefore \quad \log _{ a }{ y } =q$

So…

$LHS\\ \\ =\ln { \left( { a }^{ m }\cdot { a }^{ q } \right) } \\ \\ =\ln { \left( { a }^{ \left( m+q \right) } \right) } \\ \\ =\left( m+q \right) \cdot \ln { a } \\ \\ =\left( \log _{ a }{ x } +\log _{ a }{ y } \right) \cdot \log _{ e }{ a } \\ \\ =\frac { \log _{ a }{ x } +\log _{ a }{ y } }{ \log _{ a }{ e } } \\ \\ =\frac { \log _{ a }{ x } }{ \log _{ a }{ e } } +\frac { \log _{ a }{ y } }{ \log _{ a }{ e } } \\ \\ =\log _{ e }{ x } +\log _{ e }{ y } \\ \\ =\ln { x } +\ln { y } \\ \\ =RHS$

# Experiments With Surds & Exponentials…

$\sqrt { a } \sqrt { a } ={ a }^{ \frac { 1 }{ 2 } }{ a }^{ \frac { 1 }{ 2 } }={ a }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }=a\\ \\ \sqrt [ 3 ]{ a } \sqrt [ 3 ]{ a } \sqrt [ 3 ]{ a } ={ a }^{ \frac { 1 }{ 3 } }{ a }^{ \frac { 1 }{ 3 } }{ a }^{ \frac { 1 }{ 3 } }={ a }^{ \frac { 1 }{ 3 } +\frac { 1 }{ 3 } +\frac { 1 }{ 3 } }=a\\ \\ \sqrt [ 4 ]{ a } \sqrt [ 4 ]{ a } \sqrt [ 4 ]{ a } \sqrt [ 4 ]{ a } ={ a }^{ \frac { 1 }{ 4 } }{ a }^{ \frac { 1 }{ 4 } }{ a }^{ \frac { 1 }{ 4 } }{ a }^{ \frac { 1 }{ 4 } }={ a }^{ \frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } }=a$

$\sqrt { \frac { a }{ b } } \sqrt { \frac { a }{ b } } ={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 2 } }{ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 2 } }={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }=\frac { a }{ b } \\ \\ \sqrt [ 3 ]{ \frac { a }{ b } } \sqrt [ 3 ]{ \frac { a }{ b } } \sqrt [ 3 ]{ \frac { a }{ b } } ={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } }{ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } }{ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } }={ \left( \frac { a }{ b } \right) }^{ \frac { 1 }{ 3 } +\frac { 1 }{ 3 } +\frac { 1 }{ 3 } }=\frac { a }{ b }$

${ a }^{ \frac { 1 }{ 2 } }=b\\ \\ \therefore \quad a={ b }^{ 2 }\\ \\ { a }^{ \frac { 3 }{ 4 } }=b\\ \\ \therefore \quad a={ b }^{ \frac { 4 }{ 3 } }\\ \\ { a }^{ \sqrt { m } }=b\\ \\ \therefore \quad a={ b }^{ \frac { 1 }{ \sqrt { m } } }\\ \\ \therefore \quad a={ b }^{ \frac { \sqrt { m } }{ m } }\\ \\ { a }^{ \frac { 1 }{ m+\sqrt { n } } }=b\\ \\ \therefore \quad a={ b }^{ m+\sqrt { n } }$

# Surd Problem…

Simplify:

$\sqrt { 1 } \sqrt { 2 } \sqrt { 3 } \sqrt { 4 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 9 } \sqrt { 10 }$

———-

$\sqrt { 1 } \sqrt { 2 } \sqrt { 3 } \sqrt { 4 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 9 } \sqrt { 10 } \\ \\ =\sqrt { 1 } \sqrt { 4 } \sqrt { 9 } \sqrt { 2 } \sqrt { 3 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 10 } \\ \\ =1\cdot 2\cdot 3\sqrt { 2 } \sqrt { 3 } \sqrt { 5 } \sqrt { 6 } \sqrt { 7 } \sqrt { 8 } \sqrt { 10 } \\ \\ =6\sqrt { 2 } \sqrt { 3 } \sqrt { 5 } \left( \sqrt { 2 } \sqrt { 3 } \right) \sqrt { 7 } \left( \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \right) \left( \sqrt { 2 } \sqrt { 5 } \right) \\ \\ =6\sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 2 } \sqrt { 3 } \sqrt { 3 } \sqrt { 5 } \sqrt { 5 } \sqrt { 7 } \\ \\ =6\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\sqrt { 7 } \\ \\ =720\sqrt { 7 }$

# New Logarithmic Proof – Quicker Version

Prove that:

$\log _{ a }{ x } =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } }$

————-

PROOF:

$RHS\\ \\ =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } } \\ \\ =\log _{ b }{ x } \cdot \log _{ a }{ b } \\ \\ =\log _{ a }{ \left( { b }^{ \log _{ b }{ x } } \right) } \\ \\ =\log _{ a }{ x } \\ \\ =LHS$

# More Trig Shortcuts

Shortcut 1:

$\cot { x } =\frac { 1 }{ \tan { x } } \\ \\ \frac { \tan { x } }{ \cot { x } } \cdot \cot { x } =\frac { 1 }{ \tan { x } } \cdot \frac { \tan { x } }{ \cot { x } } \\ \\ \tan { x } =\frac { 1 }{ \cot { x } }$

Shortcut 2:

$\cot { x } =\frac { \cos { x } }{ \sin { x } } \\ \\ \frac { \sin { x } }{ \cos { x } \cot { x } } \cdot \cot { x } =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin { x } }{ \cos { x } \cot { x } } \\ \\ \frac { \sin { x } }{ \cos { x } } =\frac { 1 }{ \cot { x } }$

Shortcut 3:

$\sec { x } =\frac { 1 }{ \cos { x } } \\ \\ \frac { \cos { x } }{ \sec { x } } \cdot \sec { x } =\frac { 1 }{ \cos { x } } \cdot \frac { \cos { x } }{ \sec { x } } \\ \\ \cos { x } =\frac { 1 }{ \sec { x } }$

Shortcut 4:

$\csc { x } =\frac { 1 }{ \sin { x } } \\ \\ \frac { \sin { x } }{ \csc { x } } \cdot \csc { x } =\frac { 1 }{ \sin { x } } \cdot \frac { \sin { x } }{ \csc { x } } \\ \\ \sin { x } =\frac { 1 }{ \csc { x } }$

# Sine Rule Derivation In Record Time…

$\frac { 1 }{ 2 } bc\sin { A } =\frac { 1 }{ 2 } ac\sin { B } \\ \\ \frac { 2 }{ c\sin { A\sin { B } } } \cdot \frac { 1 }{ 2 } bc\sin { A } =\frac { 2 }{ c\sin { A\sin { B } } } \cdot \frac { 1 }{ 2 } ac\sin { B } \\ \\ \frac { 2bc\sin { A } }{ 2c\sin { A\sin { B } } } =\frac { 2ac\sin { B } }{ 2c\sin { A\sin { B } } } \\ \\ \frac { 2 }{ 2 } \cdot \frac { c }{ c } \cdot \frac { \sin { A } }{ \sin { A } } \cdot \frac { b }{ \sin { B } } =\frac { 2 }{ 2 } \cdot \frac { c }{ c } \cdot \frac { \sin { B } }{ \sin { B } } \cdot \frac { a }{ \sin { A } } \\ \\ \frac { b }{ \sin { B } } =\frac { a }{ \sin { A } }$

Now:

$\sin { C } =\sin { \left( 90-B+\left( 90-A \right) \right) } \\ \\ \sin { C } =\sin { \left( 180-\left( A+B \right) \right) } \\ \\ \sin { C } =\sin { 180 } \cos { \left( A+B \right) -\cos { 180 } } \sin { \left( A+B \right) } \\ \\ \sin { C } =-\left( -1 \right) \sin { \left( A+B \right) } \\ \\ \sin { C= } \sin { \left( A+B \right) } \\ \\ \sin { C } =\sin { A } \cos { B } +\cos { A } \sin { B } \\ \\ \sin { C= } \frac { 2A }{ bc } \cdot \frac { x }{ a } +\frac { \left( c-x \right) }{ b } \cdot \frac { 2A }{ ac } \\ \\ \sin { C } =\frac { 2Ax }{ acb } +\frac { 2A\left( c-x \right) }{ acb }$

$\\ \\ \sin { C } =\frac { 2Ax+2A\left( c-x \right) }{ acb } \\ \\ \sin { C } =\frac { 2A\left\{ x+\left( c-x \right) \right\} }{ acb } \\ \\ \sin { C= } \frac { 2Ac }{ acb } \\ \\ \frac { ab }{ 2 } \cdot \sin { C } =\frac { 2A }{ ab } \cdot \frac { ab }{ 2 } \\ \\ \frac { 1 }{ 2 } ab\sin { C } =A\\ \\ \therefore \quad \frac { 1 }{ 2 } ab\sin { C } =\frac { 1 }{ 2 } bc\sin { A } \\ \\ \frac { 2 }{ b\sin { A\sin { C } } } \cdot \frac { 1 }{ 2 } ab\sin { C } =\frac { 2 }{ b\sin { A\sin { C } } } \cdot \frac { 1 }{ 2 } bc\sin { A } \\ \\ \frac { a }{ \sin { A } } =\frac { c }{ \sin { C } } \\ \\ \therefore \quad \frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } }$