Derivative Check: Xe^{lnX^2}' = X^2e^{lnX^2}+2Xe^{lnX^2}

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Homework Help Overview

The discussion revolves around the differentiation of the expression \( Xe^{\ln(X^2)} \), focusing on the application of derivative rules and simplifications involving logarithmic identities.

Discussion Character

  • Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to differentiate the expression using the product rule and chain rule, while others suggest that recognizing the simplification \( e^{\ln(X^2)} = X^2 \) could lead to a more straightforward solution.

Discussion Status

Participants are exploring different methods of differentiation, with some acknowledging the potential for simplification. There is a recognition of the correctness of the original poster's approach, but also an indication that an alternative method may be simpler.

Contextual Notes

There appears to be an emphasis on finding the most efficient method for differentiation, with participants engaging in a back-and-forth regarding the correctness of the approaches taken.

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[tex]\frac{d}{dx} Xe^{lnX^2} = e^{lnX^2}+Xe^{lnX^2} \frac{1}{X^2} 2x[/tex]
[tex]=\frac{X^2 e^{lnX^2}+Xe^{lnX^2}2X}{X^2}[/tex]

is that correct?
 
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True enough, but since [tex]e^{ln(x^{2})}=x^{2}[/tex] there is an easier way to solve this..
 
oh wow...

[tex]3x^2[/tex]

is that it?
 
Right..:wink:
 

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