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How do you find the second derivative of # ln(x^2+4)# ?

Answer 1

Aurora Alvarado

#(d^2ln(x^2 + 4))/dx^2 = (8 - 2x^2)/(x^2 + 4)^2#

The chain rule is:

#(d{f(u(x))})/dx = (df(u))/(du)((du)/dx)#

Let #u(x) = x^2 + 4#, then #(df(u))/(du) =(dln(u))/(du) = 1/u# and #(du)/dx = 2x#

#(dln(x^2 + 4))/dx = (2x)/(x^2 + 4)#

#(d^2ln(x^2 + 4))/dx^2 = (d((2x)/(x^2 + 4)))/dx#

#(d((2x)/(x^2 + 4)))/dx =#

#{2(x^2 + 4) - 2x(2x)}/(x^2 + 4)^2 = #

#(8 - 2x^2)/(x^2 + 4)^2#

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Answer 2

Paisley Johnson

To find the second derivative of ( \ln(x^2+4) ), you first find the first derivative using the chain rule:

[ \frac{d}{dx}(\ln(x^2+4)) = \frac{1}{x^2+4} \cdot \frac{d}{dx}(x^2+4) = \frac{2x}{x^2+4} ]

Then, differentiate the first derivative with respect to ( x ) to find the second derivative:

[ \frac{d^2}{dx^2}(\ln(x^2+4)) = \frac{d}{dx}\left(\frac{2x}{x^2+4}\right) = \frac{2(x^2+4) - 2x(2x)}{(x^2+4)^2} ]

[ \frac{d^2}{dx^2}(\ln(x^2+4)) = \frac{2(x^2+4) - 4x^2}{(x^2+4)^2} ]

[ \frac{d^2}{dx^2}(\ln(x^2+4)) = \frac{8}{(x^2+4)^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.