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

Answer 1

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

# f(x)=ln(x^2+2) #

Using the chain rule we have: # f'(x)=1/(x^2+2) *(2x)# # :. f'(x)=(2x)/(x^2+2) #

To find the second derivative we now need to use the product rule. Remember: # d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2 #; so

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

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Answer 2

Cameron Alexander

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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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.