# How do you find the second derivative of #f(x)=ln (x^2+2)#?

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To find the second derivative of ( f(x) = \ln(x^2 + 2) ), first, find the first derivative using the chain rule. The first derivative 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, differentiate ( f'(x) ) using the quotient rule to find the second derivative:

[ f''(x) = \frac{d}{dx} \left(\frac{2x}{x^2 + 2}\right) = \frac{(2)(x^2 + 2) - 2x(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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