Logarithmic Derivative of y = 5ln(7lnx)

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Homework Help Overview

The discussion revolves around finding the derivative of the function y = 5ln(7lnx), focusing on the application of logarithmic differentiation and the chain rule.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to differentiate the function using various approaches, questioning the application of the chain rule and the meaning of certain terms in their equations. There is confusion regarding the notation used and the steps taken in the differentiation process.

Discussion Status

Some participants have provided alternative expressions for the derivative, while others have pointed out potential errors in the initial attempts. There is an ongoing exploration of the correct application of differentiation rules without a clear consensus on the final form of the derivative.

Contextual Notes

Participants are grappling with notation and terminology, such as "lnx0," which may not be standard. The discussion reflects the challenges of applying calculus concepts correctly in a homework context.

Jan Hill
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Homework Statement



y = 5ln(7lnx)

Homework Equations



y' = 5 x 1/7lnx +97lnx0 x 1/x

= 5(7lnx)/7lnx + 5ln x 7/x


The Attempt at a Solution



y' = 5 + 35ln/x

is this right?
 
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Jan Hill said:

Homework Statement



y = 5ln(7lnx)

Homework Equations



y' = 5 x 1/7lnx +97lnx0 x 1/x

= 5(7lnx)/7lnx + 5ln x 7/x
You're completely forgetting the chain rule.
Where did the 97 come from?
What does lnx0 mean?

y' = 5 * 1/(7 lnx) * d/dx(7 lnx) = ?
Jan Hill said:

The Attempt at a Solution



y' = 5 + 35ln/x

is this right?
 
then I guess it should be

y' = 5/(7lnx) x 7/x

Y' = 35/x(7lnx)

Is that right?
 
Jan Hill said:
then I guess it should be

y' = 5/(7lnx) x 7/x

Y' = 35/x(7lnx)

Is that right?
Yes, but it should be simplified to 5/(x ln(x))
 

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