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Decart Group TitleBest ResponseYou've already chosen the best response.0
what do you have so far
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Angela, are you allowed to use a table for derivatives of inverse trig functions? If not, there is a way we can do this using a triangle, it just takes a few more steps than normal.
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
I honestly don't have a clue of wat to do :( @Decart @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Answer the question silly billy >:O Are you given access to a table of Inverse Trig Derivatives? :)
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
no I don't think so:) @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\Large\bf y=arcsec 2x\qquad\qquad\to\qquad\qquad 2x=\sec y\]Understand what I did so far? I just rewrote the relationship so we don't have that inverse function anymore.
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
that's possible?? :o and yes I understand:)
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So what we want to do is some sneaky triangle math that you might remember back from Trig. \[\Large\bf \sec y\quad=\quad 2x\quad=\quad \frac{2x}{1}\quad=\quad \frac{hypotenuse}{adjacent}\]We want to draw this relationship on a triangle.
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1384737357099:dwUnderstand how we'll label this?
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
lol let me look at it a lil more xp
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
wait, does this apply if its f(x)=arc sec2x ??:) @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
You'll see how it connects later on D:
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
ok lol go on:) @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1384737660359:dwSo we'll label our sides according to our trig identity.
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We can find the missing side using the Pythagorean Theorem, yes?
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1384737872951:dw
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So we've successfully setup a triangle that relates back to secant.
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
NOW let's take a derivative.\[\Large\bf \sec y\quad=\quad 2x\]Taking our derivative with respect to x, what do you get? :x
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
opposite side= sqrt of 4x^21 ?
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
2 ?? @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
2 ?? water you talking about? :u
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
the derivative ?? @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
that gives us the derivative of the `right side`, yes :) 2x turns into 2. What about the left side?
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
sec y tan y ??@zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Mmm close!! Don't forget your chain rule! We're taking the derivative of a ything with respect to x. So the chain rule should also give us a y'
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\Large \bf \sec y \tan y \cdot y'\quad=\quad 2\]
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
oooh! @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \bf \sec y \tan y \cdot \color{#DD4747 }{y'}\quad=\quad 2\]Remember what this y represented at the start of the problem? \(\Large y=\sec 2x\) So this y' here represents the derivative of sec2x, the thing we've been looking for all along! :O
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So we just need to isolate the pretty red y'
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
and write it in terms of x* which we can do using the triangle.
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So if we divide some stuff over:\[\Large \bf y'\quad=\quad \frac{2}{\sec y \tan y}\]Then we'll recall our identities to rewrite this as:\[\Large\bf y'\quad=\quad 2\cos y \cot y\]
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Then we use our triangle to get our final answer:\[\Large\bf y'\quad=\quad 2\frac{adjacent}{hypotenuse}\cdot \frac{adjacent}{opposite}\]
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
and that's it?? @zepdrix
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So when you plug those sides in you'll get a big ole mess that you want to simplify :o
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Err a little mess :)
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Plugging in the sides gives us: \[\Large\bf y'\quad=\quad 2\left(\frac{1}{2x}\right)\left(\frac{1}{\sqrt{4x^21}}\right)\] Mmm something like that.
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
If that process was wayyyy confusing, maybe just use a table of derivatives for now:\[\Large\bf y=\sec x\qquad\qquad\to\qquad\qquad y'\quad=\quad \frac{1}{x \sqrt{x^21}}\]
 11 months ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Since we had 2x,\[\Large\bf y=\sec \color{#DD4747}{2x}\qquad\to\qquad y'\quad=\quad \frac{1}{\color{#DD4747}{2x} \sqrt{(\color{#DD4747}{2x})^21}}(\color{#DD4747}{2x})'\]
 11 months ago

angelarios Group TitleBest ResponseYou've already chosen the best response.0
and just solve?? @zepdrix
 11 months ago
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