Question

Prove that d/dc arcsecu=1/u sqrt u^2-1 du/dx

Answer 1

Two things, the first, and more important being... what's the derivative of the arcsecant function?

Answer 2

\[\huge \frac{d}{dx}\sec^{-1}x\]

Answer 3

yup.

Answer 4

To understand this, let's look at this function...
\[\huge y = \sec^{-1}x\]

Answer 5

okie

Answer 6

Now, this is (pretty much) the same as
\[\huge \sec \ y = x\]
Agreed?

Answer 7

i agree, :)

Answer 8

Well, then, use implicit differentiation, and find dy/dx.

Answer 9

Are you still with me?

Answer 10

yes lol just trying to remember how to do implicits.

Answer 11

Okay. Keep me posted :)

Answer 12

how do i start it again? lol

Answer 13

You differentiate x as normal.
You differentiate y as you would with x, but when you're done, multiply it by dy/dx

Answer 14

im confused lol.

Answer 15

x would just be 1 then?

Answer 16

That's right. As for y, you treat it as a function of x.

Answer 17

And you apply the chain rule. Differentiate y as if it were an x, and afterwards, when the dust settles, multiply the differentiated expression by dy/dx, or y'

Answer 18

so i got sec y * y' = 1?

Answer 19

sec y?
I said differentiate it first, before you multiply the y' :P

Answer 20

so secytany * y'?

Answer 21

Bingo :)
Let me write that down for you...
After differentiating both sides with respect to x... we get
\[\huge (\sec \ y \ \tan \ y)\frac{dy}{dx}=1\]
Ok? :P

Answer 22

yup!

Answer 23

Now, solve for dy/dx.
What I mean is that make it so that dy/dx is alone on one side of the equation.

Answer 24

dy/dx=1/secytany

Answer 25

Very good :)
\[\huge \frac{dy}{dx}=\frac{1}{\sec \ y \ \tan \ y }\]

Answer 26

ok

Answer 27

Now, earlier, we started with
\[\huge y = \sec^{-1}x\]
right?

Answer 28

oui

Answer 29

Well, that's the same as
\[\huge \sec \ y = x\]
yes?

Answer 30

yuppp

Answer 31

Well, then, put that to one side, and let's get to work on this...
\[\large \frac{dy}{dx}=\frac1{\sec \ y \ \tan \ y}\]

Remember the pythagorean identities? Specifically the one that relates secant to tangent?

Answer 32

sec=h/o?

Answer 33

t=o/a

Answer 34

No... this identity...
\[\large \cos^2\theta+\sin^2\theta=1\]
Remember it?

Answer 35

oya

Answer 36

Well, related to it is the identity involving the tangent and the secant function. Remember it?

Answer 37

yes

Answer 38

What is it?

Answer 39

sec^2=1+tan^2

Answer 40

Okay, very good :)
\[\large \sec^2x=1+\tan^2x\]
So it follows that
\[\huge \sec^2x-1=\tan^2x\]
catch me so far?

Answer 41

yup.

Answer 42

Ok, we take the square root of both sides, we get...
\[\huge \sqrt{\sec^2x-1}=\tan \ x\]
Got it?

Answer 43

yupp!

Answer 44

So, back to this...
\[\huge \frac{dy}{dx}=\frac1{\sec \ y \ \tan \ y}\]
Due to the conclusion we just arrived at...
We can replace tan y with...
\[\huge \frac{dy}{dx}=\frac1{(\sec \ y)\sqrt{\sec^2y-1}}\]
Getting it?

Answer 45

now i just sub back in x because secy=x

Answer 46

You catch on really quickly. So you're finally left with...?

Answer 47

dy/dx=1/xsqrtx^2-1

Answer 48

Et voila...
Well done.
With this, you should have no trouble at all answering your question, just one application of chain rule, and you're all set :D

Answer 49

so to apply the fchain rule, i should just write the dy/dx at the end?

Answer 50

Not dy/dx but du/dx

Answer 51

yeah thats where im a little confused, how to make it du/dx because thaqts what the question asks for.

Answer 52

Well, could you restate your question? I suspect a typo in your post...

Answer 53

there is also a second part to this question:
use the formula we just came up with in a to find the derivative of y=arcsec(x/a) where a is a constant.

Answer 54

d/dx arcsecu=1/u squrt u^2-1 du/dx

Answer 55

Chain rule. You just get the derivative of the arcsecant, except replace x with x/a
And then multiply it to the derivative of x/a, which is just...?

Answer 56

i have to run to class!
ill be back in about an hr or so.
thanks for the help and explanation, it really helped me a lot. : )

Answer 57

back!