Derivative of Hyperbolic Trigonometric Function

In summary, the student attempted to solve a homework equation, but was not able to do so using the product rule. They were able to solve it using the quotient rule and then simplified the answer. Half credit was given for messy notation and making the problem longer.
  • #1
gurtaj
20
0

Homework Statement



derivative of (e^x-e^-x)/(e^x+e^-x)


Homework Equations





The Attempt at a Solution



attachment.php?attachmentid=35933&stc=1&d=1306367699.jpg

This answer was marked 2/4 on the test, can anyone help me get the right answer please ?
 

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  • #2
Hi gurtaj! :smile:

I don't understand your first step: you have [itex](e^x-e^{-x})(e^x+e^{-x})^{-1}[/itex] and you wish to derive it. The product rule says that this equals

[tex](e^x-e^{-x})^\prime(e^x+e^{-x})^{-1}+(e^x-e^{-x})((e^x+e^{-x})^{-1})^\prime[/tex]

You did the second part of the sum allright, but you didn't differentiate anything in the first part of the sum...
 
Last edited:
  • #3
oh sorry the next step after
[itex](e^x-e^{-x})(e^x+e^{-x})^{-1}[/itex]
I meant to put
[itex](e^x+e^{-x}) (e^x+e^{-x})^{-1} + (e^x-e^{-x})(-1)(e^x+e^{-x})^{-2}(e^x-e^{-x})[/itex]

so i did
derivative of the first one times second , then derivative of the second one times first
 
  • #4
OK, so that expression will evaluate easily to the answer...
 
  • #5
i think that is right too but somehow it was marked 2/4, and the teacher wrote expand and simplify
 
  • #6
Yes, the final answer is correct. But if I was your teacher I would have given 2/4 too. The point is that your notation is really messy and you can't really see what you're doing :frown: And you're making the problem longer than it needs to be...
 
  • #7
Quotient rule will greatly simplify your answer, and be much cleaner.
 
  • #8
micromass said:
Yes, the final answer is correct. But if I was your teacher I would have given 2/4 too. The point is that your notation is really messy and you can't really see what you're doing :frown: And you're making the problem longer than it needs to be...

The teacher didn't cut 2 marks for messy notations or making the problem longer, she said its the answer that's wrong. She said you can expand that answer even more
 
  • #9
I tried Quotient rule and end up with same answer

[itex](e^x-e^{-x})/(e^x+e^{-x})[/itex]

[itex](e^x+e^{-x}) (e^x+e^{-x})-(e^x-e^{-x})(e^x-e^{-x})/ (e^x+e^{-x})^{2}[/itex]

[itex] 1-((e^x-e^{-x})^{2}/ (e^x+e^{-x})^{2}) [/itex]
 
  • #10
Then I guess your teacher wants you to simplify your answer. Put your fractions on one denominator and work out the squares...
A bit silly to subtract points for that, unless she stated that your answer must be simplified...
 
  • #11
gurtaj said:
She said you can expand that answer even more

Ohhhh, I see what's going on here.

Hint: Look and sinh and cosh.

You still the problem right though. I can't see why your teacher would give you half credit.

(Also, look at tanh)
 
  • #12
gb7nash said:
Ohhhh, I see what's going on here.

Hint: Look and sinh and cosh.

You still the problem right though. I can't see why your teacher would give you half credit.

(Also, look at tanh)

We never learned Hyperbolic function, so she wasn't expecting that from us
 
  • #13
I have no idea then. I think your instructor's crazy.
 
  • #14
gb7nash said:
I have no idea then. I think your instructor's crazy.

I second that! :smile:
 

What is a derivative in calculus?

A derivative in calculus is a measure of the rate at which a function changes. It represents the slope of a tangent line to a curve at a given point and is calculated using the limit concept. In simpler terms, it shows how much a function is changing at a specific point.

How do you find the derivative of a function?

To find the derivative of a function, you need to use the rules of differentiation. These rules include the power rule, product rule, quotient rule, and chain rule. You can also use the limit definition of a derivative to find the derivative of a function.

Why is the derivative important in calculus?

The derivative is important in calculus because it has many practical applications in fields such as physics, engineering, economics, and more. It is used to solve optimization problems, find maximum and minimum values, and analyze the behavior of a function at a specific point.

What is the difference between the derivative and the rate of change?

The derivative and the rate of change are closely related concepts, but there is a subtle difference between them. The derivative is a general concept that applies to functions, while the rate of change is specific to a particular variable. In other words, the derivative measures the overall rate of change of a function, while the rate of change measures the change of a specific variable within that function.

Can you have a derivative at a point where a function is not defined?

No, you cannot have a derivative at a point where a function is not defined. The definition of a derivative requires the function to be continuous at the point where you are trying to find the derivative. If the function is not defined at that point, it cannot be continuous and therefore cannot have a derivative.

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