Differention (could you check my solutions please)

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Homework Help Overview

The discussion revolves around differentiating the function y = e^(-x)sin(3x). Participants are examining different methods for differentiation, specifically the product rule and logarithmic differentiation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to differentiate using both the product rule and logarithmic differentiation, presenting their results for review. Some participants question the validity of the logarithmic approach, particularly the transition from ln(e^(-x)) to 1/x, and discuss the derivative of the logarithmic function.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing feedback on the correctness of the methods used. There is an ongoing exploration of the logarithmic differentiation approach, with some participants offering clarifications and corrections without reaching a consensus.

Contextual Notes

There is a focus on understanding the properties of logarithms and their derivatives, with some confusion regarding the application of these concepts in the context of differentiation. Participants are also navigating the implications of their assumptions about the logarithmic function.

enosthapa
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Homework Statement



Differentiate y=e-xsin(3x)

2.The attempt at a solution

With Product rule
y'=-e-xSin(3x)+3e-xCos(3x)

And Logarithm way
Taking log both sides

lny= lne-x+ln(Sin3x)

lny= 1/x+ln(Sin3x)

1/y*dy/dx=-x-2+3(Cos3x/Sin3x)

y'=-(e-xSin3x)/x2+3e-xCos3x
 
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your product rule one looks good , but I don't think log one is .
why do you go from ln(e^-x) to 1/x
dervative of ln(e^(-x)) is -1
 
cragar said:
your product rule one looks good , but I don't think log one is .
why do you go from ln(e^-x) to 1/x
dervative of ln(e^(-x)) is -1

I would make sense but this is what i did

ln(e^-x) = 1/(lne^x)
=1/(x.lne)
=1/x
 
ok but you see the difference ln(e^-x) = ln(1/(e^x))
 
Oh yes thankx... but can I do
ln(e^-x) = -x lne =-x
 
enosthapa said:
Oh yes thankx... but can I do
ln(e^-x) = -x lne =-x

Of course!

ln(t) is the inverse function of e(t), so ln(et ) = t .
 

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