Derivatives of ln including Absolute Value

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SUMMARY

The discussion focuses on finding the derivative of the function f(x) = ln|(x+2)/(x³ - 1)|. The domain is established as {x ∈ ℝ, x ≠ 1, -2} due to the restrictions imposed by the logarithm. The derivative is calculated using the chain rule, resulting in f'(x) = 1/(x+2) - 1/(x-1) - (2x+1)/(x²). It is confirmed that no separate derivatives are needed for positive and negative values due to the properties of the natural logarithm involving absolute values.

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  • Understanding of logarithmic functions and their properties
  • Familiarity with the chain rule in calculus
  • Knowledge of domain restrictions for logarithmic expressions
  • Ability to factor polynomial expressions
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  • Study the application of the chain rule in calculus
  • Explore the properties of logarithmic functions, particularly with absolute values
  • Learn how to determine the domain of composite functions
  • Practice finding derivatives of more complex logarithmic functions
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Students studying calculus, particularly those focusing on derivatives of logarithmic functions, and educators seeking to clarify concepts related to absolute values in calculus.

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Homework Statement


Determine the domain and find the derivative

f(x) = ln|(x+2)/(x3 - 1)|


Homework Equations





The Attempt at a Solution



We can factor x3 - 1 = (x-1)(x2) + x + 1)
From this we know that x ≠ 1 or -2 because 1 would be undefined at -2 would cause the function to be ln 0 which is not possible.
Domain = {x\inℝ, x ≠ 1,-2}
f(x) = ln(x+2) - ln(x3 - 1)
= ln(x+2) - ln[(x-1)(x2) + x + 1)]
= ln(x+2) - ln(x-1) - ln(x2) + x + 1)
f'(x) = 1/(x+2) - 1/(x-1) -(2x+1)/(x2) + x + 1)

This is where i stopped but i know that there is an absolute value so I'm not sure if i have to find when x is positive and when x is negative and then find the separate derivatives. Please help me figure out if up to here is fine or if i have to keep going and if i do have to keep going, how would i go about finding the derivative? Thank you.
 
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Nope, no separate derivatives needed. I don't remember the proof, but if its a natural log involving an absolute value, the derivative would be the same for f(x) if abs() wasn't there.

Do you want me to find it?

Also.. why are you factoring things?

Just use the chain rule.

d/dx (ln|w|) = 1/w * dw/du * du/dx

feel free to consult http://www.wolframalpha.com/input/?i=d/dx+(f(x)+=+ln|(x+2)/(x3+-+1)|)

if you need further help...
 
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