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angelarios
find the derivative of the function. arc sec2x
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.
I honestly don't have a clue of wat to do :( @Decart @zepdrix
Answer the question silly billy >:O Are you given access to a table of Inverse Trig Derivatives? :)
no I don't think so:) @zepdrix
\[\Large\bf y=arcsec 2x\qquad\qquad\to\qquad\qquad 2x=\sec y\]Understand what I did so far? I just rewrote the relationship so we don't have that inverse function anymore.
that's possible?? :o and yes I understand:)
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.
|dw:1384737357099:dw|Understand how we'll label this?
lol let me look at it a lil more xp
wait, does this apply if its f(x)=arc sec2x ??:) @zepdrix
You'll see how it connects later on D:
ok lol go on:) @zepdrix
|dw:1384737660359:dw|So we'll label our sides according to our trig identity.
We can find the missing side using the Pythagorean Theorem, yes?
So we've successfully setup a triangle that relates back to secant.
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
opposite side= sqrt of 4x^2-1 ?
2 ?? water you talking about? :u
the derivative ?? @zepdrix
that gives us the derivative of the `right side`, yes :) 2x turns into 2. What about the left side?
sec y tan y ??@zepdrix
Mmm close!! Don't forget your chain rule! We're taking the derivative of a y-thing with respect to x. So the chain rule should also give us a y'
\[\Large \bf \sec y \tan y \cdot y'\quad=\quad 2\]
\[\huge \bf \sec y \tan y \cdot \color{#DD4747 }{y'}\quad=\quad 2\]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
So we just need to isolate the pretty red y'
and write it in terms of x* which we can do using the triangle.
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:\[\Large\bf y'\quad=\quad 2\cos y \cot y\]
Then we use our triangle to get our final answer:\[\Large\bf y'\quad=\quad 2\frac{adjacent}{hypotenuse}\cdot \frac{adjacent}{opposite}\]
and that's it?? @zepdrix
So when you plug those sides in you'll get a big ole mess that you want to simplify :o
Plugging in the sides gives us: \[\Large\bf y'\quad=\quad 2\left(\frac{1}{2x}\right)\left(\frac{1}{\sqrt{4x^2-1}}\right)\] Mmm something like that.
If that process was wayyyy confusing, maybe just use a table of derivatives for now:\[\Large\bf y=\sec x\qquad\qquad\to\qquad\qquad y'\quad=\quad \frac{1}{x \sqrt{x^2-1}}\]
Since we had 2x,\[\Large\bf y=\sec \color{#DD4747}{2x}\qquad\to\qquad y'\quad=\quad \frac{1}{\color{#DD4747}{2x} \sqrt{(\color{#DD4747}{2x})^2-1}}(\color{#DD4747}{2x})'\]
and just solve?? @zepdrix