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

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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,

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

How do you find the derivative of f(x) = ln (3x^2 - 1)f(x)=ln(3x2−1)f(x) = ln (3x^2 - 1)?