How Do You Apply the Chain Rule to Differentiate \( y = xe^{-x^2} \)?

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Homework Help Overview

The discussion revolves around differentiating the function \( y = xe^{-x^2} \), focusing on the application of the chain rule and product rule in calculus.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to apply both the product rule and the chain rule for differentiation. Some express uncertainty about how to start, while others suggest using specific rules for derivatives of exponential functions.

Discussion Status

Several participants have provided guidance on using the product rule and chain rule, indicating a collaborative effort to clarify the differentiation process. There is acknowledgment of multiple interpretations of the problem, but no explicit consensus has been reached.

Contextual Notes

Some participants mention a desire for initial guidance without providing complete solutions, reflecting the forum's emphasis on learning through discussion.

physicsed
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[SOLVED] another chain rule: easy one

y=xe^{-x^2}

i have no i dea how to start.
f'= x^{x^2} or -2x^blah blah blah

just get me started and i'll promise you i will finish it myself
 
Last edited:
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This is actually an application of the product rule, then the chain rule.
 
physicsed said:
y=xe^{-x^2}

i have no i dea how to start.
f'= x^{x^2} or -2x^blah blah blah

just get me started and i'll promise you i will finish it myself

You will probably want to use the product rule and the rule for finding the derivative of e^{g(x)}
 
Product rule!

f(x)=e^{-x^2}

derivative of e is itself, times the derivative of it's exponent.
 
wow 3 replies all at 22:49 ... you just got the royal treatment :D
 
rocophysics said:
wow 3 replies all at 22:49 ... you just got the royal treatment :D

ROFL:smile:
 
Start with the product rule. When you get to needing to find d/dx(e^(-x^2)) then remember the chain rule says (f(g(x)))'=f'(g(x))*g'(x). f is exp. g(x)=-x^2. So?
 
Solved it
thanks
 
Geez. I'm really late.
 
  • #10
Y'= e^{-x2}(1-2x^{2})
thanks for the help
 
  • #11
Looks right^^
 

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