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What is the second derivative of #f(x)=ln (x^2+2)#?

Answer 1

Emily Ammerman

You can use the Chain Rule to get: #f'(x)=1/(x^2+2)*2x=(2x)/(x^2+2)#

Then the Quotient Rule to get:

#f''(x)=(2(x^2+2)-2x*2x)/(x^2+2)^2=# #=(2x^2+4-4x^2)/(x^2+2)^2=(4-2x^2)/(x^2+2)^2#

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Answer 2

Charles Anctil

To find the second derivative of ( f(x) = \ln(x^2 + 2) ), we first find the first derivative and then differentiate it again.

Given ( f(x) = \ln(x^2 + 2) ),

The first derivative, ( f'(x) ), is: [ f'(x) = \frac{d}{dx}(\ln(x^2 + 2)) ] [ = \frac{1}{x^2 + 2} \cdot \frac{d}{dx}(x^2 + 2) ] [ = \frac{2x}{x^2 + 2} ]

Now, to find the second derivative, ( f''(x) ): [ f''(x) = \frac{d}{dx}\left(\frac{2x}{x^2 + 2}\right) ] [ = \frac{(x^2 + 2) \cdot \frac{d}{dx}(2x) - 2x \cdot \frac{d}{dx}(x^2 + 2)}{(x^2 + 2)^2} ] [ = \frac{(x^2 + 2) \cdot 2 - 2x \cdot 2x}{(x^2 + 2)^2} ] [ = \frac{2x^2 + 4 - 4x^2}{(x^2 + 2)^2} ] [ = \frac{-2x^2 + 4}{(x^2 + 2)^2} ]

So, the second derivative of ( f(x) = \ln(x^2 + 2) ) is: [ f''(x) = \frac{-2x^2 + 4}{(x^2 + 2)^2} ]

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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.