Second Derivative of Exponential Function

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Homework Help Overview

The discussion revolves around finding the second derivative of exponential functions, specifically e^{ax} and e^{-ax}. The subject area pertains to calculus and differentiation, particularly focusing on the application of the chain rule in the context of exponential functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion regarding the application of the chain rule to exponential functions and questions whether they are overcomplicating the problem. Some participants discuss the differentiation process, with one participant outlining the use of the chain rule and another confirming the steps taken to find the second derivative.

Discussion Status

The discussion is ongoing, with participants providing insights into the differentiation process. There is an attempt to clarify the application of the chain rule, and one participant has shared their calculations for the second derivative, seeking confirmation of their correctness.

Contextual Notes

There is mention of a textbook that lacks clarity on the chain rule's application to exponential functions, which may be contributing to the original poster's confusion. The assumption that "a" is a constant is also noted in the discussion.

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Homework Statement



Find the second derivative of:

[tex]e^{ax}[/tex]

and

[tex]e^{-ax}[/tex]

Homework Equations





The Attempt at a Solution



The book that I am using seems to have been very vague on how to take the derivatives of exponential functions. I am aware that:

[tex]\frac {d(e^{x})} {dx} = e^{x}[/tex]

but it says literally nothing about how the chain rule applies to exponential function, or does it? Am I just making it too difficult? Please help!
 
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[tex]\frac {d(e^{u})} {dx} = e^{u} \frac{du}{dx}[/tex]
 
Assuming "a" as a constant, you can consider two functions.

One is [tex]f(y)=e^y[/tex] and the other [tex]y=ax[/tex], so the derivative in respect to x would be:

[tex]\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}=...[/tex].

You just need to calculate the differentials tand to the same again to get the result.
 
Last edited:
Ok so I tried the method listed, I am hoping someone can confirm that this is correct:

[tex]y = e^{ax}[/tex]

[tex]y' = a(e^{ax})[/tex]

[tex]y'' = a^2(e^{ax})[/tex]

Thanks!
 
That's it ;)
 

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