# Arcsine Question

Discussion in 'Math' started by ElectronicsFanatic, Apr 9, 2012.

1. ### ElectronicsFanatic Thread Starter Member

Feb 12, 2012
30
1
Ok I just got done taking a test and I wasn't sure the answer to the following question. Maybe someone can help me understand this better. This is all new material to me so I am having some trouble getting it down.

What is wrong with a problem that requires finding the derivative y=arcsin(x²+1)

I know the formula for this is $\frac{1}{\sqrt{1-u^2$

I don't understand why this wont work? Does anyone have any hints to send my way. I really don't know why it wouldn't work

2. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Well, if you set x^2 + 1 equal to 0 and try to solve for x, you're going to get an imaginary number. You can't take the arcsin of an imaginary number. That is if I'm remembering my factoring correctly

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3. ### amilton542 Active Member

Nov 13, 2010
494
64
It's a composite function.

You need to use the chain-rule: dy/dx = [(dy/du)(du/dx)]

The change in u cancels.

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4. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Why are you taking the derivative? It's a simple arcsin question (sin^-1). It's to find the angle of a specific value on a coordinate plane. Nothing more.

5. ### amilton542 Active Member

Nov 13, 2010
494
64
If you refer to the question carefully, he said, "What is wrong with a problem that requires finding the derivative y=arcsin(x²+1)."

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Feb 20, 2011
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7. ### studiot AAC Fanatic!

Nov 9, 2007
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consider

x² >= 0 thus

(x²+1) >= 1
and only equal if x=0.

But sin (θ) <= 1

so the only value of x that you can take this arc sin is x=0

You cannot take the differential of a function that has a single point - it does not vary.

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8. ### amilton542 Active Member

Nov 13, 2010
494
64
I don't believe it. Thanks for pointing that out, I missed that one. I need to observe a function more carefully before approaching it, thanks Studiot!

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9. ### ElectronicsFanatic Thread Starter Member

Feb 12, 2012
30
1
Ok, I think I get what you guys are all saying, so, I am going to go through this process using the formula I have learned and anyone let me know when I make a mistake.

$\frac{2x}{1}\times\frac{1}{\sqrt{1-(x^2+1)^2}}$

$\frac{2x}{\sqrt{1-(x^2+1)^2}$

This is where I start to not quite understand

I think the final answer would be the following:

$\frac{2}{\sqrt{-1(x^2+1)}}$

When I set $x^2+1=0$ I understand how I am getting an imaginary number.

That would be $x=\sqrt{-1}$

So what happened to the square on $(x^2+1)^2$ to just become $(x^2+1)$

I am probably making this harder than it needs to be. I really understand what you guys are all saying I am trying to relate it to the formula. Thanks everyone for the answers. I am understanding it a whole lot better than i did before.

10. ### amilton542 Active Member

Nov 13, 2010
494
64
Look at post #7.

11. ### ElectronicsFanatic Thread Starter Member

Feb 12, 2012
30
1
ok I applied the chain rule to $(x^2+1)^2$ as you stated amilton542. I can now see how the answer is applied.

That took a little longer than i wanted but i understand now thanks everyone. I think Studiot made it the easiest to understand. I am still trying to remember to apply easier methods to the problems. You guys are all a wealth of information. I wont forget that one again.

Last edited: Apr 9, 2012
12. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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I think you were directed to my post#7.

y=arcsin(x²+1)

Has no derivative.

That is dy/dx does not exist anywhere.

Oh and welcome to All About Circuits.

13. ### amilton542 Active Member

Nov 13, 2010
494
64
Nooooooo, the functions fixed its not changing thats the problem.

I would like to apologise for giving you the chain-rule, I read your question to fast.

Refer to what studiot has pointed-out, its post #7.

14. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
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dy/dx is usually implied when someone is told to take the derivative of a function, I think.

15. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
There is no neighbourhoods about the points x=0, y= (4n+1)∏/2, which are the only points on the graph of

y=arcsin(x²+1)

The function does not exist for any other value of x.
It is not a question of imaginary numbers, x is not imaginary.

Last edited: Apr 9, 2012
16. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
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Yeah, that was from when I misunderstood the question. I misread it and thought it was just asking to solve it, not take the derivative.

17. ### studiot AAC Fanatic!

Nov 9, 2007
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So does our new member now understand?

18. ### ElectronicsFanatic Thread Starter Member

Feb 12, 2012
30
1
Yes. Thanks everyone. The farthest I got in math was precalculus about 12 years ago in high school. This is all new material for me so I appreciate everyone being patient with me and all the help.

I am slowly trying to remember everything I have learned thus far. I am just about done with Calculus I and will take a break from math for a couple of months and start up again with Calculus II in June. I hope to have all my math finished by December so I can then focus on all the funner parts of electronics. Not that I don't like math, because I enjoy trying to figure things out using math. It just isn't always fun.