## angelarios one year ago find the derivative of the function. arc sec2x

1. Decart

what do you have so far

2. zepdrix

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.

3. angelarios

I honestly don't have a clue of wat to do :( @Decart @zepdrix

4. zepdrix

Answer the question silly billy >:O Are you given access to a table of Inverse Trig Derivatives? :)

5. angelarios

no I don't think so:) @zepdrix

6. 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.

7. angelarios

that's possible?? :o and yes I understand:)

8. zepdrix

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.

9. zepdrix

|dw:1384737357099:dw|Understand how we'll label this?

10. angelarios

lol let me look at it a lil more xp

11. angelarios

wait, does this apply if its f(x)=arc sec2x ??:) @zepdrix

12. zepdrix

You'll see how it connects later on D:

13. angelarios

ok lol go on:) @zepdrix

14. zepdrix

|dw:1384737660359:dw|So we'll label our sides according to our trig identity.

15. zepdrix

We can find the missing side using the Pythagorean Theorem, yes?

16. angelarios

yes:)

17. zepdrix

|dw:1384737872951:dw|

18. zepdrix

So we've successfully setup a triangle that relates back to secant.

19. zepdrix

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

20. angelarios

opposite side= sqrt of 4x^2-1 ?

21. angelarios

2 ?? @zepdrix

22. zepdrix

2 ?? water you talking about? :u

23. angelarios

the derivative ?? @zepdrix

24. zepdrix

that gives us the derivative of the right side, yes :) 2x turns into 2. What about the left side?

25. angelarios

sec y tan y ??@zepdrix

26. 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'

27. zepdrix

$\Large \bf \sec y \tan y \cdot y'\quad=\quad 2$

28. angelarios

oooh! @zepdrix

29. zepdrix

$\huge \bf \sec y \tan y \cdot \color{#DD4747 }{y'}\quad=\quad 2$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

30. zepdrix

So we just need to isolate the pretty red y'

31. zepdrix

and write it in terms of x* which we can do using the triangle.

32. zepdrix

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:$\Large\bf y'\quad=\quad 2\cos y \cot y$

33. zepdrix

Then we use our triangle to get our final answer:$\Large\bf y'\quad=\quad 2\frac{adjacent}{hypotenuse}\cdot \frac{adjacent}{opposite}$

34. angelarios

and that's it?? @zepdrix

35. zepdrix

So when you plug those sides in you'll get a big ole mess that you want to simplify :o

36. zepdrix

Err a little mess :)

37. zepdrix

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.

38. zepdrix

If that process was wayyyy confusing, maybe just use a table of derivatives for now:$\Large\bf y=\sec x\qquad\qquad\to\qquad\qquad y'\quad=\quad \frac{1}{x \sqrt{x^2-1}}$

39. zepdrix

Since we had 2x,$\Large\bf y=\sec \color{#DD4747}{2x}\qquad\to\qquad y'\quad=\quad \frac{1}{\color{#DD4747}{2x} \sqrt{(\color{#DD4747}{2x})^2-1}}(\color{#DD4747}{2x})'$

40. angelarios

and just solve?? @zepdrix