Struggling with Derivatives and Simplifying Trigonometric Functions

[jdepalma]

Homework Statement

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

Homework Equations

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

Find dX/db

The Attempt at a Solution

Too much to put here

 

[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)

 

[Cyosis]

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

 

[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)"

Thanks in advance!

 

[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 ?

 

[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.

 

[Cyosis]

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

 

[rootX]

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.

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)

 

[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!

 

[Cyosis]

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

<br /> X=2 \; \text{arccsc}\sqrt{1+\alpha^2 b^2}<br />

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.

 

[guerom00]

My bad… Indeed, there's no Arccsc in the answer

 

[Dick]

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

<br /> X=2 \; \text{arccsc}\sqrt{1+\alpha^2 b^2}<br />

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.

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

 

[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?

 

[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).

<br /> \frac{d}{dx} \text{arccsc(x)}=\frac{-1}{x \sqrt{x^2-1}}<br />

 

[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 argument 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!

 

[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'

 

[Dick]

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 argument 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!

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.

 

[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!

 

[Cyosis]

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

 

[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!