Derivative of Hyperbolic function!

  • Thread starter Rizzamabob
  • Start date
  • #1
21
0
Hey,
Need help in the steps to take to find the derivative of
y=cosh(x^2 + loge(x))
and e^y +ytanh(x) = x

I have never seen them before, therefor not sure which rule to use, im thinking the second needs partial derivatives :bugeye:
Thanks!!
 

Answers and Replies

  • #2
86
0
Look up 'hyperbolic functions' in the index of your calculus book.

In my calculus book the rules for differentiating and integrating hyperbolic functions are buried in a chapter, rather than listed on the front inside cover.

I bet your book is set up in the same way.
 
  • #3
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,549
8
And one other thing...

Rizzamabob said:
y=cosh(x^2 + loge(x))

Is the part in red supposed to be [itex]\log_e(x)[/itex] (as in [itex]\ln(x)[/itex]), or is that supposed to be [itex]\log\left(e^x\right)[/itex]?
 
  • #4
21
0
Left my book at uni, thats why i need help now :(
 
  • #5
21
0
Tom Mattson said:
And one other thing...



Is the part in red supposed to be [itex]\log_e(x)[/itex] (as in [itex]\ln(x)[/itex]), or is that supposed to be [itex]\log\left(e^x\right)[/itex]?
[itex]\log_e(x)[/itex]
Thanks:bugeye:
 
  • #6
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,549
8
Rizzamabob said:
Left my book at uni, thats why i need help now :(

Google, buddy, Google.

We're happy to help you with your questions, but as per the notice at the top of this Forum, you have to make some attempt. I'll introduce you to your two new best friends:

Wikipedia
Math World

You will find definitions to all things mathematical at those sites. So, please search for "hyperbolic functions", compose your thoughts on the problem, and let us know where you're stuck. You'll be better off for it!
 
  • #7
21
0
Thanks, atm im working this way

y = (x^2 (e^x + e^-x))/2 + ([itex]\log_e(x)[/itex](e^x + e^-x))/2

===>

y = ((e^x + e^-x) (x^2 + [itex]\log_e(x)[/itex]))/2

:surprised
Then try to find dy/dx i have a feeling it will be YUCK :surprised
 
Last edited:
  • #8
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,549
8
Rizzamabob said:
Thanks, atm im working this way

OK, now we're cooking.

I think that it would behoove you to either work out or look up the derivative of the hyperbolic cosine function. You'll need it later.

y = (x^2 (e^x + e^-x))/2 + ([itex]\log_e(x)[/itex](e^x + e^-x))/2

No, that's not it.

The way you've written the problem, it should be the hyperbolic cosine of [itex]x^2+\log_e(x)[/tex], but above you've interpreted it as [itex]\cosh(x)[/itex] times that function.

It should be as follows:

[tex]\cosh\left(x^2+log_e(x)\right)=\frac{e^{\left(x^2+log_e(x)\right)}+e^{-\left(x^2+log_e(x)\right)}}{2}[/tex]

Now, that does look nasty but fortunately you don't have to deal with it as it is written on the right hand side. Once you know the derivative of the hyperbolic cosine, you need only to recognize that you have the following composite function:

[tex]f(u)=\cosh(u)[/tex]
[tex]u(x)=x^2+\log_e(x)[/tex]

Do you know what rule you need to use to differentiate a composite function?
 
Last edited:

Related Threads on Derivative of Hyperbolic function!

Replies
2
Views
4K
Replies
4
Views
22K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
890
Replies
4
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
6
Views
1K
Top