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angelarios
 2 years ago
find the derivative of the function.
arc sec2x
angelarios
 2 years ago
find the derivative of the function. arc sec2x

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0that's possible?? :o and yes I understand:)

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

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

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

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

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

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0opposite side= sqrt of 4x^21 ?

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0sec y tan y ??@zepdrix

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

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

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0and that's it?? @zepdrix

zepdrix
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0and just solve?? @zepdrix
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