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How do you find the second derivative of # ln(x^2+5)# ?

Answer 1

Emily Ammerman

#(-2(x^2-5))/(x^2+5)^2#

Use natural logarithm derivative rules

#dy/dx =(2x)/(x^2+5) #

For second derivative use the quotient rule

#(d^2y)/(dx^2)= (-2(x^2-5))/(x^2+5)^2#

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Answer 2

Emilia Aguilar

To find the second derivative of ( \ln(x^2 + 5) ), follow these steps:

Find the first derivative using the chain rule: [ \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} ] [ u = x^2 + 5 ] [ \frac{du}{dx} = 2x ] [ \frac{d}{dx} \ln(x^2 + 5) = \frac{1}{x^2 + 5} \cdot 2x ] [ \frac{d}{dx} \ln(x^2 + 5) = \frac{2x}{x^2 + 5} ]
Find the second derivative using the quotient rule: [ \frac{d^2}{dx^2} \ln(u) = \frac{d}{dx} \left( \frac{1}{u} \cdot \frac{du}{dx} \right) ] [ = \frac{d}{dx} \left( \frac{2x}{x^2 + 5} \right) ] [ = \frac{(2)(x^2 + 5) - 2x(2x)}{(x^2 + 5)^2} ] [ = \frac{2x^2 + 10 - 4x^2}{(x^2 + 5)^2} ] [ = \frac{-2x^2 + 10}{(x^2 + 5)^2} ] [ = \frac{-2(x^2 - 5)}{(x^2 + 5)^2} ]

Therefore, the second derivative of ( \ln(x^2 + 5) ) is ( \frac{-2(x^2 - 5)}{(x^2 + 5)^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.