How to Differentiate y=2e(2x+1) Using the Chain Rule?

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Discussion Overview

The discussion focuses on differentiating the function y=2e(2x+1) using the chain rule, exploring the application of differentiation techniques in calculus.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant seeks clarification on how to differentiate y=2e(2x+1) using the chain rule, initially letting u= (2x+1) and finding du/dx = 2.
  • Another participant points out the presence of multiplication and exponentiation in the expression y= 2eU.
  • A participant questions whether a second application of the chain rule is necessary and proposes that dy/dx = 4e(2x+1).
  • One participant clarifies the interpretation of the expression, suggesting that it should be written as y= 2e^(2x+1) and explains that the chain rule is applied only once, leading to dy/dx = 4e^(2x+1).
  • Another participant expresses gratitude for the clarification provided.

Areas of Agreement / Disagreement

There is some confusion regarding the correct interpretation of the expression and the application of the chain rule. While some participants agree on the final derivative, the initial interpretation of the function remains contested.

Contextual Notes

Participants have not fully resolved the ambiguity in the notation used for the function, which affects the differentiation process.

v_pino
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Hi everyone,

I'm new to this forum... I hope I've posted in the right section...

How do I differentiate y=2e(2x+1) using the chain rule?

I let u= (2x+1)

so du/dx = 2

but how do I differentiate y= 2eU ?

thank you :)
 
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v_pino said:
how do I differentiate y= 2eU ?

thank you :)
Well, I see a multiplication and an exponentiation in that expression...
 
Does it mean that I should use the chain rule again?

and I get dy/du = 2eU

so dy/dx = 4e(2x+1) ?
 
Last edited:
The fact that a problem uses the word "differentiate" does not mean it is a differential equation! I am moving this to the Calculus and Analysis forum.
 
First, do you mean
[tex]y= 2e^{2x+1}[/itex]<br /> which in "ASCII" would be y= 2e^(2x+1). What you wrote, I would interpret as 2e <b>times</b>(2x+1) and there is no need for the chain rule!<br /> <br /> If you let u= 2x+1, then, yes, the chain rule says that the derivative of y= 2e^u, with respect to x, is dy/dx= 2e^u (du/dx). Since you have already determined that du/dx= 2, that is dy/dx= 2e^(2x+1) (2)= 4 e^(2x+1). There is no need for a second application of the chain rule.[/tex]
 
thank you very much :)
 

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