Derivitive with ln - don't understand

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SUMMARY

The discussion focuses on differentiating the function y = ln(2x^2 - 2) / (x^2 - 1). The derivative is established as y' = -ln(2x^2 - 2)(2x)(x^2 - 1)^-2 + 4x / (2x^2 - 2)(x^2 - 1)^-1. The confusion arises around the term 4x / (2x^2 - 2), which is clarified through the application of the chain rule in differentiation, specifically using the formula for the derivative of ln(f(x)).

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Homework Statement



Differentiate the following:
y = ln(2x^2 - 2) / x^2 - 1



The Attempt at a Solution



The answer is y' = -ln(2x^2 - 2)(2x)(x^2 - 1)^-2 + 4x / 2x^2 - 2 (x^2 - 1)^-1

I understand all of it except 4x / 2x^2 - 2, how does ln(2x^2 - 2) derive into 4x / 2x^2 - 2?
 
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\frac{d}{dx}(ln(f(x))=\frac{f'(x)}{f(x)}
 
Oh okay, thanks, never seen that one before.
 
zeion said:
Oh okay, thanks, never seen that one before.
If you look closely, you'll notice that's just the chain rule...
 
Really? How?
 
To see how it is the chain rule, think about what the derivative of ln x is.
 

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