# Homework Help: Help with a nasty derivative

1. Apr 24, 2010

### jdepalma

1. The problem statement, all variables and given/known data
Hey all, I'm looking for an idea on how to solve this derivative. I've tried all the rules I can think and I've gotten nowhere. Help is appreciated

2. Relevant equations
X = 2*arccosecant(1+((2*b*E)/B)^2)^(1/2)

Find dX/db

3. The attempt at a solution

Too much to put here

2. Apr 24, 2010

### guerom00

Hi

Do you know the derivative of arccos(x) ? If yes, there is no particular difficulties although it might be lengthy… You basically have a function of a function of a function of arccos(x)

3. Apr 24, 2010

### Cyosis

Do you know what the derivative of arccsc(x) with respect to x is?

4. Apr 24, 2010

### jdepalma

Yes, the derivative of arccsc(x) is:

d/dx arccsc(x) = -1/((x)*(1-x^2)^(1/2))

And I'm not sure what you mean by "Do you know the derivative of arccos(x) ? If yes, there is no particular difficulties although it might be lengthy… You basically have a function of a function of a function of arccos(x)"

5. Apr 24, 2010

### guerom00

Well, you don't have arccsc(b) but arccsc(f(b)) where f(b)=1+((2*b*E)/B)^2 (not quite sure on what is applied the square root in your OP…)
Do you know how to derive f(g(x)) wrt x ?

6. Apr 24, 2010

### jdepalma

Ok, I will restate the original problem so its more clear.

X = 2*(arccsc(1+((2bE/B)^2)))^(1/2)

The root is of the entire arccsc term. I am still not sure where you're going with f(g(x)) wrt x. I'm not great in math and am having a hard time with this. Are you saying to compose the 2 functions?

I'm still a bit lost.

7. Apr 24, 2010

### Cyosis

Finding the derivative of the arccsc is the hard part,which you have already done. Now just apply the chainrule.

8. Apr 24, 2010

### rootX

Further simplified:
x = 2*(g)^(1/2)
g = arccsc(y)
y= 1+((2bE/B)^2)

if you want to have dx/dy given above, refer to the chain rule (as already said by all others)

9. Apr 24, 2010

### jdepalma

Ok. I tried the chain rule as all of you diagrammed above. Here's what I came up with:

dX/db = (1/(arccsc(1+((2bE/B)^2)))^(1/2))*(1/(|1+((2bE/B)^2)|*(1-(1+((2bE/B)^2))^2))^(1/2))

I am not sure if I did it right, but I took my time and made sure I did a good job. Any feedback is appreciated. Thanks so much!

10. Apr 24, 2010

### Cyosis

I am afraid that is not correct. First of all there can't be an arccsc in the final answer. Lets first tidy up the expression a little bit. All those constants make it terribly messy.

$$X=2 \; \text{arccsc}\sqrt{1+\alpha^2 b^2}$$

With $\alpha=(2E/B)$. Now take the derivative of the arccsc with respect to its argument (post this). Then take the derivative of the square root with respect to its argument (post this). Then take the derivative of the argument of the square root with respect to b (post this). We can stitch them together after that.

Last edited: Apr 24, 2010
11. Apr 24, 2010

### guerom00

12. Apr 24, 2010

### Dick

I think the problem is supposed to be
$$X=2 \; \sqrt {\text{arccsc}({1+\alpha^2 b^2})}$$
so there is an arccsc in the answer.

13. Apr 24, 2010

### jdepalma

Yes, Dick is correct. That is how the problem is supposed to read. I've tried many ways to get that derivative. I thought my chain rule expansion was correct, but maybe I didn't take the derivative of the function inside of the the arccsc.

I think I have to take the derivative of the inside of the arccsc first, then use the chain rule. Any thoughts?

14. Apr 24, 2010

### Cyosis

Looking at his answer again I think you're right. In that case jdepalma you're pretty close to the answer. You still need to apply the chain rule to the argument of the arccsc.

edit:

You made a small mistake with respect to the derivative of arccsc(x).

$$\frac{d}{dx} \text{arccsc(x)}=\frac{-1}{x \sqrt{x^2-1}}$$

Last edited: Apr 24, 2010
15. Apr 24, 2010

### jdepalma

OK! After using all of the wonderful advice I've received, here is what I came up with. Hopefully we'll be able to stitch this all together.

Using the function posted by Dick, I used the chain rule to take the derivative of the arguement of the big radical sign, setting u=arccsc(1+a^2*b^2). I got:

f'(b) = -1/((arccsc(1+a^2*b^2)*(1+a^2*b^2)*((1+a^2*b^2)^2-1)^(1/2))

I then took the derivative of the argument of the arccsc with respect to b and got:

f'(b) = (8*b*E^2)/B^2

Does this look right? I am no sure how to combine these into a usable answer. Any advice is appreciated. Thanks!!

16. Apr 24, 2010

### Cyosis

It looks pretty good, but you're missing the last term. You need to use the chain rule one more time. Substitute v=1+a^2b^2 when you're differentiating the arccsc.

f'(b) = -1/((arccsc(1+a^2*b^2)*(1+a^2*b^2)*((1+a^2*b^2)^2-1)^(1/2)) v'

17. Apr 24, 2010

### Dick

Good idea. Put u=arccsc(1+a^2*b^2). Then your function is X=2*sqrt(u). So the derivative is 2*(1/2)*u^(-1/2)*du/db. Now put v=1+a^2*b^2. u=arccsc(v) where v=1+a^2*b^2. So du/db=(du/dv)*(dv/db). du/dv=(-1)/(v*sqrt(v^2-1)). dv/db=2*a^2*b. Now just put it altogether. No single part of the differentiation is hard. It's just that the final expression is lengthy.

18. Apr 24, 2010

### jdepalma

OK. So my final answer is:

f'(b) = -1/((arccsc(1+a^2*b^2)*(1+a^2*b^2)*((1+a^2*b^2)^2-1)^(1/2)) * ((8*b*E^2)/B^2)

I think that's what you meant, as I took the derivavtive of the v you gave wrt b.

Also, while I'm here, another part of this question is to simplify the following:

csc(2*arccosecant(1+((2*b*E)/B)^2)^(1/2))

the csc of my original function X that Dick kindly put into a more reader friendly format. I tried it and got stuck with the square root. Any suggestions? Thanks!!!

19. Apr 24, 2010

### Cyosis

Looks right to me. I can't say I can see a way to simplify that second expression.

20. Apr 24, 2010

### jdepalma

Thanks for all the help. It is much appreciated.

Also, while I'm here, another part of this question is to simplify the following:

csc(2*arccosecant(1+((2*b*E)/B)^2)^(1/2))

the csc of my original function X that Dick kindly put into a more reader friendly format. I tried it and got stuck with the square root. Any suggestions? Thanks!!!