Comments on: Differentiation & Integration: A Hence Question http://www.exampaper.com.sg/questions/a-maths/differentiation-integration-a-hence-question Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: sham sihabdeen http://www.exampaper.com.sg/questions/a-maths/differentiation-integration-a-hence-question#comment-10948 sham sihabdeen Tue, 22 Jul 2008 07:49:00 +0000 http://www.exampaper.com.sg/questions/a-maths/differentiation-integration-a-hence-question#comment-10948 well explined and worth well explined and worth

]]> By: fggffggf http://www.exampaper.com.sg/questions/a-maths/differentiation-integration-a-hence-question#comment-9529 fggffggf Sat, 14 Jun 2008 16:09:16 +0000 http://www.exampaper.com.sg/questions/a-maths/differentiation-integration-a-hence-question#comment-9529 [pmath]{d/dx}{ln(cos 2x)}= (1/{cos 2x}) (- sin 2x) (2)[/pmath] [pmath]{}= {-2sin 2x}/{cos 2x}=-2tan 2x[/pmath] <span class="highlight">Yup remember your <m>{d/dx}(ln u) = {1/u}{du/dx}</m> here!</span> .'.[pmath]tan 2x={-1/2}{d/dx}{ln(cos 2x)}={d/dx}({-1/2}{ln(cos 2x)})[/pmath] <span class="highlight">This is the key part of this <em>hence</em> question: expressing tan 2<var>x</var> as a differential using your earlier answer. May Heaven help you if you're trying to integrate tan 2<var>x</var> on your own! (At least if you're an O-Level student anyway :P)</span> .'.[pmath]int{0}{pi/6}{tan 2x} dx = delim{[}{{-1/2}{ln(cos 2x)}}{]}^{pi/6}_0[/pmath] =[pmath]{-1/2}{ln(cos (2*{pi/6}))}+1/2{ln(cos (2*0))}[/pmath] =[pmath]-1/2{ln(1/2)}+1/2{ln(1)}[/pmath] =[pmath]-1/2{ln(1/2)}=0.34657[/pmath] (corr. to 5 sig. fig.) <span class="highlight">CORRECTO!</span> A little thought on the integral sign: it looks more like the figure just under the strings of a violin, but I do agree that it is rather like a tangent curve. My E+A-math teacher says tangent curves are like "pretty ladies in a beauty pageant". <span class="highlight">OMG the pageant must be held outdoors since tangent curves tend to reach <a href="/questions/a-maths/so-this-is-how-it-all-ends" rel="nofollow">high into infinity</a>! Sorry lame math jokes I know ...</span> And when I put E and A together I instantly thought of the slogan of EA games (I don't play any EA games, though): <i>EA Games - Challenge Everything.</i> Waiting for more math challenges! <span class="highlight">Stay tuned ;)</span> p.s. the math expressions are all stuck together because when I tried to separate them there was always some problems with the display. {d/dx}{ln(cos 2x)}= (1/{cos 2x}) (- sin 2x) (2)
{}= {-2sin 2x}/{cos 2x}=-2tan 2x

Yup remember your {d/dx}(ln u) = {1/u}{du/dx} here!

.'.tan 2x={-1/2}{d/dx}{ln(cos 2x)}={d/dx}({-1/2}{ln(cos 2x)})

This is the key part of this hence question: expressing tan 2x as a differential using your earlier answer. May Heaven help you if you're trying to integrate tan 2x on your own! (At least if you're an O-Level student anyway :P )

.'.int{0}{pi/6}{tan 2x} dx = delim{[}{{-1/2}{ln(cos 2x)}}{]}^{pi/6}_0

={-1/2}{ln(cos (2*{pi/6}))}+1/2{ln(cos (2*0))}

=-1/2{ln(1/2)}+1/2{ln(1)}

=-1/2{ln(1/2)}=0.34657 (corr. to 5 sig. fig.) CORRECTO!

A little thought on the integral sign: it looks more like the figure just under the strings of a violin, but I do agree that it is rather like a tangent curve. My E+A-math teacher says tangent curves are like "pretty ladies in a beauty pageant".

OMG the pageant must be held outdoors since tangent curves tend to reach high into infinity! Sorry lame math jokes I know ...

And when I put E and A together I instantly thought of the slogan of EA games (I don't play any EA games, though): EA Games - Challenge Everything.

Waiting for more math challenges! Stay tuned ;)

p.s. the math expressions are all stuck together because when I tried to separate them there was always some problems with the display.

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