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

[v_pino]

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 :)

 

[Hurkyl]

how do I differentiate y= 2eU ?

thank you :)

Well, I see a multiplication and an exponentiation in that expression...

 

[v_pino]

Does it mean that I should use the chain rule again?

and I get dy/du = 2eU

so dy/dx = 4e(2x+1) ?

 

[HallsofIvy]

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.

 

[HallsofIvy]

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]

 

[v_pino]

thank you very much :)