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Decart
 one year ago
Best ResponseYou've already chosen the best response.0what do you have so far

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Angela, 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.

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

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

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0no I don't think so:) @zepdrix

zepdrix
 one year ago
Best 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.

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0that's possible?? :o and yes I understand:)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1384737357099:dwUnderstand how we'll label this?

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0lol let me look at it a lil more xp

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You'll see how it connects later on D:

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0ok lol go on:) @zepdrix

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1384737660359:dwSo we'll label our sides according to our trig identity.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1We can find the missing side using the Pythagorean Theorem, yes?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So we've successfully setup a triangle that relates back to secant.

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

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0opposite side= sqrt of 4x^21 ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.12 ?? water you talking about? :u

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0the derivative ?? @zepdrix

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

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0sec y tan y ??@zepdrix

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Mmm 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'

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \bf \sec y \tan y \cdot y'\quad=\quad 2\]

zepdrix
 one year ago
Best 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So we just need to isolate the pretty red y'

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1and write it in terms of x* which we can do using the triangle.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So 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:\[\Large\bf y'\quad=\quad 2\cos y \cot y\]

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

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0and that's it?? @zepdrix

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Plugging 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.

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

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

angelarios
 one year ago
Best ResponseYou've already chosen the best response.0and just solve?? @zepdrix
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