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

Answer 1

Aurora Alvarado

#(dy)/dx=(12x)/(3x^2-1)#

Using chain rule, #(dy)/dx=(dy)/(du)*(du)/dx#, where

#y=2ln(u)# and #u=3x^2-1#

#(dy)/(du)=2/u#

#(du)/dx=6x#

#:.(dy)/dx=2/u*6x#

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

#=(12x)/(3x^2-1)#

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Answer 2

Ezekiel Alexander

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

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

Now, apply the derivative of the natural logarithm function:

[ = 2 \frac{1}{3x^2 - 1} \cdot \frac{d}{dx}(3x^2 - 1) ] [ = 2 \frac{1}{3x^2 - 1} \cdot 6x ] [ = \frac{12x}{3x^2 - 1} ]

So, the derivative of ( y = 2\ln(3x^2 - 1) ) is ( \frac{12x}{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.