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

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SUMMARY

The discussion focuses on applying the product rule and chain rule to differentiate the function \( y = xe^{-x^2} \). Participants emphasize the importance of starting with the product rule, followed by the chain rule for the exponential component. The derivative is confirmed as \( y' = e^{-x^2}(1 - 2x) \), demonstrating the correct application of these calculus rules. This solution highlights the interconnectedness of differentiation techniques in calculus.

PREREQUISITES
  • Understanding of the product rule in differentiation
  • Knowledge of the chain rule in calculus
  • Familiarity with exponential functions and their derivatives
  • Basic algebraic manipulation skills
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  • Study the product rule in depth with examples
  • Learn about the chain rule and its applications in differentiation
  • Explore differentiation of exponential functions, specifically \( e^{g(x)} \)
  • Practice solving complex derivatives involving multiple rules
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Students of calculus, mathematics educators, and anyone looking to strengthen their skills in differentiation techniques.

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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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