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

Answer 1

Evelyn Agard

#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)#

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Answer 2

Cameron Alexander

To find the derivative of ( f(x) = \ln(3x^2 - 1) ), you can use the chain rule. The derivative is:

[ f'(x) = \frac{1}{3x^2 - 1} \cdot \frac{d}{dx}(3x^2 - 1) ]

Applying the chain rule:

[ \frac{d}{dx}(3x^2 - 1) = 6x ]

So,

[ f'(x) = \frac{1}{3x^2 - 1} \cdot 6x = \frac{6x}{3x^2 - 1} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.