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Hyperbolic Differentiation

  • Thread starter Gondur
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  • #1
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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
 

Answers and Replies

  • #2
Curious3141
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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.
 
  • #3
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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
 
  • #4
SteamKing
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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?
 
  • #5
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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.
 
  • #6
Curious3141
Homework Helper
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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.
 
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