Is My Understanding of Basic Derivatives Correct?

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Homework Help Overview

The discussion revolves around the differentiation of exponential functions, specifically focusing on the derivatives of expressions involving \( e^x \), \( e^{-x} \), and \( e^{x+1} \). Participants are examining their understanding of basic derivative rules and the application of the chain rule.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the differentiation of various exponential functions and question the correctness of their approaches. There are discussions about the application of the chain rule and the quotient rule, as well as confusion regarding specific derivative notations.

Discussion Status

The discussion is active, with participants providing guidance on derivative rules and clarifying misunderstandings. Some participants express confusion about specific notation and the application of rules, while others attempt to clarify these points. There is no explicit consensus, but productive dialogue is ongoing.

Contextual Notes

Participants are working within the constraints of homework problems and are trying to reconcile differing answers among peers. There is mention of specific derivative rules that may not be familiar to all participants.

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[SOLVED] basic derivative question

since d/dx(e^x) = e^x

does d/dx(e^[-x]) = e^[-x]

and d/dx(e^[x+1])= e^(x+1)

the answer to one of my homework problems is different from my friends and I think that it is because I am mistaken about the stuff I just posted
 
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No; [itex]\frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)}[/itex]. Since in your first case, f(x)=1. f'(x)=1 so it doesn't matter, but it does for the second.
 
hmmm, what is wrong with this:

d/dx(e^(x+1)) = d/dx(e^[x] * e^[1]) = e^[x] * d/dx(e) + e * d/dx(e^x) = e^x * (0) + e * e^[x]= e^(x+1)
 
ggcheck said:
hmmm, what is wrong with this:

d/dx(e^(x+1)) = d/dx(e^[x] * e^[1]) = e^[x] * d/dx(e) + e * d/dx(e^x) = e^x * (0) + e * e^[x]= e^(x+1)

Nothing; that's correct. Alternatively, using the notation in my previous post, we have that f(x)=x+1, and so f'(x)=1 => d/dx(e^{x+1})=e^{x+1}
 
I thought you said that the second one isn't true?
 
I am confused; is d/dx(e^(x+1)) = e^(x+1)
 
ggcheck said:
I thought you said that the second one isn't true?

I did; the second one being d/dx(e^{-x}).
 
how can I find the derivative of e^[-x]

sorry for the confusion
 
wait, I did this to it... does this work:

e^-x = 1/e^x = e^x(0) + (1)(e^x)
 
  • #10
cristo said:
Read post #2. If you have problems understanding that, then feel free to ask anything specific. I can't help you if you don't read what I write, can I?
I read post #2, I'm not familiar with that notation... I have never seen it before.
 
  • #11
ggcheck said:
wait, I did this to it... does this work:

e^-x = 1/e^x = e^x(0) + (1)(e^x)

No, you need to use the quotient rule to differentiate quotients, something you may not have done yet.

Instead, read post #2. What is f(x) in this case?
 
  • #12
cristo said:
No, you need to use the quotient rule to differentiate quotients, something you may not have done yet.

Instead, read post #2. What is f(x) in this case?
f(x) = -x

right?
 
  • #13
ggcheck said:
f(x) = -x

right?

Correct. So, what's f'(x)? Note that the prime here just means "derivative wrt x" so, f'(x)=d/dx(f(x))=d/dx(-x)

[NB: I deleted my post a few above, the one you quoted, as it seemed a little abrupt. Sorry about that.]
 
  • #14
which is -1?
 
  • #15
ggcheck said:
which is -1?

Yup, and so, using the formula in #2, what is the derivative of e^{-x}?
 
  • #16
-e^{-x} ?
 
  • #17
ggcheck said:
-e^{-x} ?

Correct!
 
  • #18
ugh, I am screwing up somewhere here...

e^{-x} = 1/(e^x) if I apply the quotient rule... [e^x * (0) - (1) * e^x] / (e^x)^2

-(e^x)/(e^{2x})

EDIT: switched signs
 
Last edited:
  • #19
nm, after I cancel I get the same thing...

thank you very much for your help
 
  • #20
would you mind taking a look at the problem that started all of this madness?
 
  • #21
d/dx[e^(x+1) + e^x]/(e+1)

the answer that I got was e^x
 
  • #22
ggcheck said:
d/dx[e^(x+1) + e^x]/(e+1)

the answer that I got was e^x

Well, this is equal to [tex]\frac{1}{e+1}\frac{d}{dx}(e^{x+1}+e^x})[/tex]. From above, we know that d/dx(e^x)=e^x, and d/dx(e^{x+1})=e^{x+1}, so we obtain [tex]\frac{e^x+e^{x+1}}{1+e}=\frac{e^x(1+e)}{1+e}=e^x[/tex]
 
  • #23
Thanks a lot, btw what software are you using to type that out?
 
  • #24
It's latex that's installed on the forum. If you click on advanced reply (or quote a post) then click on the little [itex]\Sigma[/itex] icon on the toolbar you will have a drop down menu to use. Alternatively, simply click on some of the maths to view the input commands.
 
  • #25
Thanks again.
 
  • #26
ggcheck said:
Thanks again.

You're welcome!
 

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