How Do You Differentiate y=1/(cosh x) Using the Chain Rule?

Click For Summary

Homework Help Overview

The discussion revolves around differentiating the function y = 1/(cosh x) using the chain rule. Participants are exploring the application of differentiation techniques and the properties of hyperbolic functions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants suggest starting with the definition of cosh(x) in terms of exponentials and rewriting the function. Others discuss the use of the chain rule and question the necessity of substituting variables.

Discussion Status

There are various approaches being explored, including the use of the chain rule and the potential simplification by letting u = cosh x. Participants are questioning assumptions about derivatives of hyperbolic functions and discussing the clarity of their steps.

Contextual Notes

Some participants express frustration with the clarity of their posts and the formatting of images, which may affect the readability of the discussion. There is also mention of the need for efficiency in methods used for differentiation.

Gondur
Messages
25
Reaction score
0

Homework Statement



y=1 / (cos h x), find dy/dx


Homework Equations



chain rule and coshx=(e^x+e^-x)/2

The Attempt at a Solution



IdRxUyK.jpg
 
Physics news on Phys.org
Resize your image! Way too big.

The most obvious error is ##\frac{dy}{du} = \ln |u|##. Sure that's the derivative, and not the integral?

Other than that, for Chain Rule, it's generally not helpful to substitute variables like this. Not wrong, but it can overcomplicate things.
 
Curious3141 said:
Resize your image! Way too big.

The most obvious error is ##\frac{dy}{du} = \ln |u|##. Sure that's the derivative, and not the integral?

Other than that, for Chain Rule, it's generally not helpful to substitute variables like this. Not wrong, but it can overcomplicate things.

What an idiot I am. This mistake is proof that I am tired and should get some sleep. Well what other method do you suggest I use, I'd definitely like to know if it's more efficient.

You can resize my image by clicking Ctrl and - on the keyboard. If you have a mouse with a wheel then turn the wheel towards you (downwards) while holding down Ctrl
 
Start with the definition of cosh(x) in terms of the exponentials and re-write 1/cosh(x). What is dy/dx when y = 1/x or x^-1?
 
SteamKing said:
Start with the definition of cosh(x) in terms of the exponentials and re-write 1/cosh(x). What is dy/dx when y = 1/x or x^-1?

I've already solved it.
 
Gondur said:
What an idiot I am. This mistake is proof that I am tired and should get some sleep. Well what other method do you suggest I use, I'd definitely like to know if it's more efficient.

Essentially what you did, but you don't have to spell out every step explicitly.

You're using ##\frac{dy}{dx} = \frac{dy}{du}\frac{du}{da}\frac{da}{dx}##, but there's no need to spell out every step.

Are you allowed to assume the derivative of the standard hyperbolic functions, like ##\frac{d}{dx}\cosh x = \sinh x##?

If so, isn't it easier to let ##u = \cosh x## and ##y = \frac{1}{u}##? Again, you shouldn't need to spell it out like this in different variables, this is just for clarity.

You can resize my image by clicking Ctrl and - on the keyboard. If you have a mouse with a wheel then turn the wheel towards you (downwards) while holding down Ctrl

Yes, I know how to resize images - which is how I viewed yours. The problem is that it becomes impossible to view your text (and mine, when I post) in the same sizing. The font becomes too small.

If you're going to post here regularly, you would be much better served by learning basic LaTex.
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K