How do you find the derivative of #f(x) = ln (3x^2 - 1)#?

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Answer 1 · 2016-06-21T18:21:17

#f'(x)=(6x)/(3x^2-1)#

There is a rule for differentiating natural logarithm functions:

If #f(x)=ln(g(x))#, then #f'(x)=(g'(x))/g(x)#.

This can be derived using the chain rule:

Since #d/dxln(x)=1/x#, we see that #d/dxln(g(x))=1/(g(x))*g'(x)=(g'(x))/g(x)#.

So, when we have #f(x)=ln(3x^2-1)#, we see that #g(x)=3x^2-1# and its derivative is #g'(x)=6x#.

Thus,

#f'(x)=(g'(x))/g(x)=(6x)/(3x^2-1)#