# Differentiate y = arccosec(x)Watch

#1
I get . Can someone else try it and tell me if I'm right?
0
10 years ago
#2
yep
0
10 years ago
#3
Yep, same as me.

Edit: bah, snap.
0
#4
Thanks, guys. I'm trying to show for myself why by differentiating them both and integrating them with limits to show that they are the same. Is this a good way of doing it?
0
10 years ago
#5
No not really. Do it anyway though, it might be instructive.

arcsin(1/x) = y
siny = 1/x
cosecy = x
y = arccosecx

EDIT: actually, you need to justify the first and last steps by looking at domains and ranges of the functions.
0
10 years ago
#6
is this stuff done in further maths or C3?

EDIT: nevermind. just did it using C3 knowledge.
0
#7
(Original post by Speleo)
No not really. Do it anyway though, it might be instructive.

arcsin(1/x) = y
siny = 1/x
cosecy = x
y = arccosecx

EDIT: actually, you need to justify the first and last steps by looking at domains and ranges of the functions.
Yeah you're totally right, it's a stupid way of doing it. Mind, after showing differentiates to give ...

0
10 years ago
#8
is this stuff done in further maths or C3?

EDIT: nevermind. just did it using C3 knowledge.
Although the actual content required to do it is in C3, its formally introduced (atleast on our exam board) in FP2.
0
10 years ago
#9
Actually that's quite a clever way to do it...

For x>0, since the functions agree at 0 you can take the integral of -1/tsqrt(t^2-1) dt between 0 and x to show that arccosecx = arcsin(1/x) and similarly for x<0.
0
10 years ago
#10
(Original post by Speleo)
Actually that's quite a clever way to do it...

For x>0, since the functions agree at 0 you can take the integral of -1/tsqrt(t^2-1) dt between 0 and x to show that arccosecx = arcsin(1/x) and similarly for x<0.
Ah I was just going to say that; using a general case rather than the specific one Zii used.
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#11
Edit: Does the fact that there's a discontinuity at zero not generate a big problem in my working?
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10 years ago
#12
I didn't actually check that 0 worked, makes no difference though, replace (literally) every instance of '0' in my last post with '1'.
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#13
I suppose I could have just integrated between 1 and 2 or something. It's just basically to show that the two potentially different constants generated are in fact equal (i.e. the answers are arcsin(1/x) + C or arccosec(x) + K - - - but C and K are in fact equal). C and K are shown to be equal by integrating within limits.
0
10 years ago
#14
Oh I see. It seems more solid to me to show that arcsin(1/x) = the definite integral dt + pi/2 = arccosec(x) for all x in the appropriate ranges but it doesn't really matter I suppose.
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#15
I don't know what you mean "the definite integral + pi/2" ?
0
10 years ago
#16
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2
0
#17
(Original post by Speleo)
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2
Why the ?

Edit: never mind, I get why now. Just scribbled it on a piece of paper .
0
10 years ago
#18
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arcsin(1/x) - arcsin(1/1) = arcsin(1/x) - pi/2
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arccosec(x) - arccosex(1) = arccosec(x) - pi/2

=>
arcsin(1/x) = [[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2 = arccosec(x)

Choosing to use a definite integral is mostly personal preference, indefinite integrals are messier.
0
10 years ago
#19
(Original post by Zii)
Why the ?

Edit: never mind, I get why now. Just scribbled it on a piece of paper .
Out of curiosity, are you just figuring this out for fun, or is it for some kind of assignment?
0
#20
(Original post by n1r4v)
Out of curiosity, are you just figuring this out for fun, or is it for some kind of assignment?
For fun.

(Original post by Speleo)
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arcsin(1/x) - arcsin(1/1) = arcsin(1/x) - pi/2
[[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] = arccosec(x) - arccosex(1) = arccosec(x) - pi/2

=>
arcsin(1/x) = [[the integral of -1/t*sqrt(t^2-1) dt between 1 and x]] + pi/2 = arccosec(x)

Choosing to use a definite integral is mostly personal preference, indefinite integrals are messier.
Yeah I've never actually done that before, wow!

I just tried it with:

OR

(after doing the necessary work...first time was a simple "derivative sitting beside it job" and the second I rewrote as )

Which means:

Cool stuff.
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