Hi all,
Hope you are well.
Well I am having a little problem with Hyperbolic Functions....
How exactly would you work the following equation out....?
f 1/3x+5 dx
Just to let you know I know that f means function and other terms in the equation, however I am unsure how you would 'evaluate the above equation.
Would you need to separate the equation into something like....
1) sinh x =
2) cosh x =
3) tanh x =
sech x =???
If so how would I do this.
I imagine the derivatives of these equations below are...
d/dx (cosh x) = sinh x
d/dx (sinh x) = coshx
Correct =/?
However even if they are I do know how to use these analogue trig functions within the hyperbolic function above =/, I am stuck....
Hope you can help + appreciate your time.
Hyperbolic Functions  Math Course, SelfStudy
Announcements  Posted on  

At uni? What do you think of the careers support?  12022016  
How proud of your uni are you? Take our survey for the chance to win a holiday!  09022016 


Thanks 
Coursework.info, I am trying my best to understand Math, it seems I need to keep trying as there is noone available.

I don't understand what you are trying to do? I assume f means to integrate the function because of the dx at the end? Please clarify what you are doing and I can try to help.
If you are integrating then surely you can just use by recognition with log base e 
Are you integrating? Is it (1/3x)+5 or 1/(3x+5). In either case, you don't need the hyperbolics.

(Original post by George1337)
I don't understand what you are trying to do? I assume f means to integrate the function because of the dx at the end? Please clarify what you are doing and I can try to help.
If you are integrating then surely you can just use by recognition with log base e(Original post by tory88)
Are you integrating? Is it (1/3x)+5 or 1/(3x+5). In either case, you don't need the hyperbolics.
It is 1/(3x+5).
Assuming there is no need for hyperbolics, how is meant to be solved, oh and it does have a 'dx' during the end.
Apologies if the equation was confusing.
I could solve it through a calculator and attempt to work out how it was done.
However, your step by step guidance to be precise would be appreciated as there are many broadened knowledge in Math on this forum.
Thank you and all the best for all,
makaveli33
Postscript: What is strange is that if it does not need 'hyperbolics' why is it on an Alison (alison.com) course (Advanced Math I) this way, the assumption is because it is a contribution and it may be wrong.
Hope you can advise.
Kind regards.
Postscript II: To use a rough estimate of calculation, would it be....
0.333333*x+5
By the way  '0.333' is recurring, thanks. 
(Original post by makaveli33)
Hi,
It is 1/(3x+5).
Assuming there is no need for hyperbolics, how is meant to be solved, oh and it does have a 'dx' during the end.

(Original post by tory88)
But are you integrating it? If so it is a simple case of f'(x)/f(x) and if you're doing hyperbolics I imagine you are capable of integrating that? 
(Original post by makaveli33)
Hi,
It is 1/(3x+5).
Assuming there is no need for hyperbolics, how is meant to be solved, oh and it does have a 'dx' during the end.
Apologies if the equation was confusing.
I could solve it through a calculator and attempt to work out how it was done.
However, your step by step guidance to be precise would be appreciated as there are many broadened knowledge in Math on this forum.
Thank you and all the best for all,
makaveli33
Postscript: What is strange is that if it does not need 'hyperbolics' why is it on an Alison (alison.com) course (Advanced Math I) this way, the assumption is because it is a contribution and it may be wrong.
Hope you can advise.
Kind regards.
Postscript II: To use a rough estimate of calculation, would it be....
0.333333*x+5
By the way  '0.333' is recurring, thanks. 
(Original post by makaveli33)
Sorry, I did not mention it is an antiderivative, :/ 
(Original post by tory88)
In which case, can you get 1/(3x+5) into the standard form f'(x)/f(x)? 
(Original post by makaveli33)
x = 3 or 3???? 
(Original post by tory88)
This is an integration (antiderivative if you prefer)  you can't solve for x like that. What, for instance, is the antiderivative of 1/x?
Antiderivative = ln(x)
I know why, because they are opposites.
Deep into the theory, the reasons are, because an antiderivative is the first step of the equation, a derivative is the second step of the equation or a solved equation as such, the antiderivative goes to the original equation based on the later one.
The problem I find with this is, having difficulty working out how to quickly and efficiently do larger sums/equations.
Thanks. 
The general rule for this sort of thing is to change it into the form of derivative over function, then it is just ln(function). So with 1/x, x is the function and 1 is its derivative. With that, how can you get 1/(3x+5) into this form?

Hello,
So you have: integrate [1/(3x+5)] dx
If you want to solve this quickly you can use a method called "Advanced guessing".
since you have something similar to 1/x, we know that 1/x dx = ln(x).
Let us say that y=ln(3x+5)
differentiating this we get I=1/(3x+5) * 3 =3/(3x+5), so I= 3/(3x+5)
However we need I to equal to 1/(3x+5), so, we must divide y by 3.
When y=(1/3)(ln(3x+5) differentiating this will give 1/(3x+5)
Hope that helps!
Let me know if this helps
Thanks! 
(Original post by sulexk)
Hello,
So you have: integrate [1/(3x+5)] dx
If you want to solve this quickly you can use a method called "Advanced guessing".
since you have something similar to 1/x, we know that 1/x dx = ln(x).
Let us say that y=ln(3x+5)
differentiating this we get I=1/(3x+5) * 3 =3/(3x+5), so I= 3/(3x+5)
However we need I to equal to 1/(3x+5), so, we must divide y by 3.
When y=(1/3)(ln(3x+5) differentiating this will give 1/(3x+5)
Hope that helps!
Let me know if this helps
Thanks!(Original post by tory88)
The general rule for this sort of thing is to change it into the form of derivative over function, then it is just ln(function). So with 1/x, x is the function and 1 is its derivative. With that, how can you get 1/(3x+5) into this form?
ln(3x+5)
Assuming this is the way.
Explained, essentially....
Because ln(x) is 1/x (opposite) then....
You times 1/x by 3 it makes 1/3x and the +5 is nonchanging as it is not a x value.
Thank you so much.
You guys are amazing.
I hope that what was said makes sense and was correct, however have a gut feeling there is understanding.
I will rep everyone on this thread for their contributions.
That advancedguessing is an amazing technique =)!!!!
Using smaller figures to make larger ones and advguessing is very useful, thank you ever so much! 
(Original post by makaveli33)
OMG, Thank you, this makes....
ln(3x+5)
Assuming this is the way.
Explained, essentially....
Because ln(x) is 1/x (opposite) then....
You times 1/x by 3 it makes 1/3x and the +5 is nonchanging as it is not a x value.
Thank you so much.
You guys are amazing.
I hope that what was said makes sense and was correct, however have a gut feeling there is understanding.
I will rep everyone on this thread for their contributions.
That advancedguessing is an amazing technique =)!!!!
Using smaller figures to make larger ones and advguessing is very useful, thank you ever so much!
You have ln(3x+5), differentiating this, will give [1/(3x+5)] *3, the *3 comes from differentiating what is inside the brackets. I can explain this using the "CHAIN RULE":
Don't Worry, if you have never seen this before, I will try and explain it as simply as I can:
Let us use the above for our example:
say "y=ln(3x+5)", and I said to you "I want to differentiate this", you will say to me "you have a function (3x+5) within another function ln(). so we have a function within a function!
Suppose, I said to you let us say that u=3x+5.
then simply put y=ln(u)
now if I want to differentiate, I can only find dy/du, since the right hand side is in terms of u, but I want to find dy/dx, so what do I do?
Well now let us use some intriguing manipulation:
(dy/du) * (du/dx) = dy/dx
now how do we find dy/du and du/dx
Well, dy/du = derivative of y=ln(u) which is 1/u
so dy/du=1/u
we know u=3x+5, so du/dx = 3
then substituting these into the above: (dy/du) * (du/dx) = dy/dx
(1/u)*(3)=3/u as u=3x+5
then dy/dx = 3/(3x+5)
and there you have it a manipulation of the chain rule to find the derivative of y=ln(3x+5)
If you want some practice with this, feel free to try some examples below:
Find the derivative of each of the following:
1) y= 1/(4x+5)
2) y= sin(3x+5)
3) y=1/(x^2 + 1)
4) y=(x+1)^2
5)y=cos(x^2 + 1)
6) y=2/sin(x)
When you finished feel free to let me know, and I'll answer them and put up the solutions!
Practice is the way forward! If you want me to provide you with some applied problems, let me know! (I can give you some rate of change problems!).
hope that helps!! 
(Original post by sulexk)
Yes.
You have ln(3x+5), differentiating this, will give [1/(3x+5)] *3, the *3 comes from differentiating what is inside the brackets. I can explain this using the "CHAIN RULE":
Don't Worry, if you have never seen this before, I will try and explain it as simply as I can:
Let us use the above for our example:
say "y=ln(3x+5)", and I said to you "I want to differentiate this", you will say to me "you have a function (3x+5) within another function ln(). so we have a function within a function!
Suppose, I said to you let us say that u=3x+5.
then simply put y=ln(u)
now if I want to differentiate, I can only find dy/du, since the right hand side is in terms of u, but I want to find dy/dx, so what do I do?
Well now let us use some intriguing manipulation:
(dy/du) * (du/dx) = dy/dx
now how do we find dy/du and du/dx
Well, dy/du = derivative of y=ln(u) which is 1/u
so dy/du=1/u
we know u=3x+5, so du/dx = 3
then substituting these into the above: (dy/du) * (du/dx) = dy/dx
(1/u)*(3)=3/u as u=3x+5
then dy/dx = 3/(3x+5)
and there you have it a manipulation of the chain rule to find the derivative of y=3x+5
If you want some practice with this, feel free to try some examples below:
Find the derivative of each of the following:
1) y= 1/(4x+5)
2) y= sin(3x+5)
3) y=1/(x^2 + 1)
4) y=(x+1)^2
5)y=cos(x^2 + 1)
6) y=2/sin(x)
When you finished feel free to let me know, and I'll answer them and put up the solutions!
Practice is the way forward! If you want me to provide you with some applied problems, let me know! (I can give you some rate of change problems!).
hope that helps!!
ln(2x+sin)
Using this method....
((dy/du) * (du/dx) = dy/dx))
then....
1/u * (2) = 2/u and u= 2x + sine or sin
Hope this is correct.
By the way, your math methods are amazing, sulexk, you make Math and calculus appear so easy by your problem solving, you deserve a tap on the back! =) 
(Original post by makaveli33)
is 6)....
ln(2x+sin)
Using this method....
((dy/du) * (du/dx) = dy/dx))
then....
1/u * (2) = 2/u and u= 2x + sine or sin
Hope this is correct.
By the way, your math methods are amazing, sulexk, you make Math and calculus appear so easy by your problem solving, you deserve a tap on the back! =)
the last one is a bit tricky, it actually doesn't require the use of ln(x):
So we have, y=2/sin(x)
if we let u=sin(x), then y=2/u=2u^1
so dy/du=2u^2
du/dx = cos(x)
put the two together: (dy/du)*(du/dx) = dy/dx = (2u^2)*(cos(x))
(2u^2)*(cos(x))=(2(sin(x)^2))*(cos(x))
(2(sin(x)^2)*(cos(x)) = [2/(sin(x)^2)]*(cos(x))=2cos(x)/[sin(x)^2]
Hope that helps!
If you need any more problems for the chain rule, or any other differentiation type problems, feel free to let me know! 
(Original post by sulexk)
Hi,
the last one is a bit tricky, it actually doesn't require the use of ln(x):
So we have, y=2/sin(x)
if we let u=sin(x), then y=2/u=2u^1
so dy/du=2u^2
du/dx = cos(x)
put the two together: (dy/du)*(du/dx) = dy/dx = (2u^2)*(cos(x))
(2u^2)*(cos(x))=(2(sin(x)^2))*(cos(x))
(2(sin(x)^2)*(cos(x)) = [2/(sin(x)^2)]*(cos(x))=2cos(x)/[sin(x)^2]
Hope that helps!
If you need any more problems for the chain rule, or any other differentiation type problems, feel free to let me know!
There are a lot of rules to remember :/
Will do some reading on the chain rule.
Thanks =). 
(Original post by sulexk)
Hi,
the last one is a bit tricky, it actually doesn't require the use of ln(x):
So we have, y=2/sin(x)
if we let u=sin(x), then y=2/u=2u^1
so dy/du=2u^2
du/dx = cos(x)
put the two together: (dy/du)*(du/dx) = dy/dx = (2u^2)*(cos(x))
(2u^2)*(cos(x))=(2(sin(x)^2))*(cos(x))
(2(sin(x)^2)*(cos(x)) = [2/(sin(x)^2)]*(cos(x))=2cos(x)/[sin(x)^2]
Hope that helps!
If you need any more problems for the chain rule, or any other differentiation type problems, feel free to let me know!
I have done some reading on the chain rule, there are many variations for the type of integer or type, using this...
http://www.sosmath.com/calculus/diff/der04/der04.html
I think it is a matter of practice, I will try to solve some of your questions as soon as possible, using the support given and the website, it is all a matter of practice =).
Thanks.
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: