Find First Derivative of 2/(e^x + e^-x)

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Homework Help Overview

The discussion revolves around finding the first derivative of the function 2/(e^x + e^-x), which falls under the subject area of calculus, specifically differentiation techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine how to approach the differentiation, considering whether to manipulate the denominator. Some participants suggest using the chain rule and inquire about its application. Others mention formulas for derivatives of products and quotients, suggesting expressing the quotient as a product.

Discussion Status

The discussion is active, with participants providing guidance on differentiation techniques, including the chain rule and product/quotient rules. There is an exploration of different methods without a clear consensus on a single approach yet.

Contextual Notes

Some participants express uncertainty about their familiarity with the chain rule and derivative formulas, indicating a potential gap in foundational knowledge that may affect their understanding of the problem.

Jason03
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Im trying to find the first derivative of the following equation:

[tex] \frac{2}{e^x+e^{-x}} [/tex]

Im trying to figure how to approach this...

i know the derivative of ex is ex...but should I move the denominator to the numerator first?
 
Last edited:
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You need to use the chain rule.
 
not to be a pest...but I am a little rusty on the chain rule...how would that apply?
 
There are formulas for the derivitave of a product and of a quotient. Are you familiar with them? If you're not familiar with the derivative of a quotient, you can express the quotient as a product:

[tex]\frac{A}{B} = A * B^{-1}[/tex]

And use your usual equation for the derivative of a product.
 
Both forms of chain rule work fine.. (product or quotient).. here's a http://www.1728.com/chainrul.htm
 
Last edited:
Cool website, Ouabache!
 
for the following problem

[tex] <br /> \int {x(5x)^{-x^2}dx}<br /> [/tex]

wouldnt this simplify to

[tex] <br /> \int{5x^{-1}dx}<br /> [/tex]
 
I don't think those two integrals are equivalent.
 
actually I wrote the first expression wrong ...this is what the first expression should be...

[tex] <br /> \int {x}{(5^{-x^2})dx}<br /> [/tex]
 
  • #10
Change the 5^(-x^2) to a power of e instead. Then use a u substitution.
 

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