Derivatives of ln including Absolute Value

In summary, the domain of the function f(x) = ln|(x+2)/(x3 - 1)| is {x∈ℝ, x ≠ 1,-2}. To find the derivative, the chain rule can be used: d/dx (ln|w|) = 1/w * dw/du * du/dx. The derivative of f(x) is 1/(x+2) - 1/(x-1) - (2x+1)/(x2) + x + 1). No separate derivatives are needed because of the absolute value. Factoring is not necessary. Wolfram Alpha can be consulted for further assistance.
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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[itex]\inℝ[/itex], 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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Related to Derivatives of ln including Absolute Value

1. What is the derivative of ln(x)?

The derivative of ln(x) is 1/x. This can be derived using the rules of logarithmic differentiation.

2. How do you find the derivative of ln|x|?

The derivative of ln|x| is 1/x for x > 0 and -1/x for x < 0. This is because the absolute value function changes the sign of negative values, resulting in a different derivative.

3. Can the derivative of ln|x| be negative?

Yes, the derivative of ln|x| can be negative since the absolute value function can result in negative values for certain inputs. This is why the derivative is split into two cases for x < 0 and x > 0.

4. How is the derivative of ln|x| related to the derivative of ln(x)?

The derivative of ln|x| is essentially the same as the derivative of ln(x), except for the negative sign for x < 0. This is because the absolute value function does not affect the derivative for positive values of x.

5. What is the chain rule for derivatives of ln including absolute value?

The chain rule for derivatives of ln including absolute value is the same as the chain rule for ln(x), with an additional factor of -1 for x < 0. This is because the absolute value function is the outer function, and the derivative of ln(x) is the inner function.

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