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Decart Group TitleBest ResponseYou've already chosen the best response.0
what do you have so far
 one year 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.
 one year 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
 one year 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? :)
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So we just need to isolate the pretty red y'
 one year 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.
 one year 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:\[\Large\bf y'\quad=\quad 2\cos y \cot y\]
 one year ago

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

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Err a little mess :)
 one year 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.
 one year 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:\[\Large\bf y=\sec x\qquad\qquad\to\qquad\qquad y'\quad=\quad \frac{1}{x \sqrt{x^21}}\]
 one year ago

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

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