# How do you find the second derivative of # ln(x/(x^2+1))# ?

*See logarithmic rules .

So far so good... Now from here, what we get is...

Thanks to the quotient rule.

If we simplify the above we get...

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To find the second derivative of ( \ln\left(\frac{x}{x^2+1}\right) ), follow these steps:

- Find the first derivative of the function.
- Find the second derivative of the resulting function.

Starting with the given function ( f(x) = \ln\left(\frac{x}{x^2+1}\right) ):

- Find the first derivative, ( f'(x) ):

[ f'(x) = \frac{d}{dx} \ln\left(\frac{x}{x^2+1}\right) ]

Using the chain rule and the derivative of natural logarithm:

[ f'(x) = \frac{1}{\frac{x}{x^2+1}} \cdot \frac{d}{dx} \left(\frac{x}{x^2+1}\right) ]

[ f'(x) = \frac{x^2 + 1}{x} \cdot \frac{(x^2+1) - x(2x)}{(x^2+1)^2} ]

[ f'(x) = \frac{x^2 + 1}{x} \cdot \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} ]

[ f'(x) = \frac{x^2 + 1}{x} \cdot \frac{1 - x^2}{(x^2+1)^2} ]

[ f'(x) = \frac{(x^2 + 1)(1 - x^2)}{x(x^2+1)^2} ]

[ f'(x) = \frac{1 - x^4}{x(x^2+1)^2} ]

- Find the second derivative, ( f''(x) ):

[ f''(x) = \frac{d}{dx} \left(\frac{1 - x^4}{x(x^2+1)^2}\right) ]

[ f''(x) = \frac{(x(x^2+1)^2)(-4x) - (1 - x^4)(x^2+1)^2 + (1 - x^4)(2x(x^2+1)(2x))}{(x(x^2+1)^2)^2} ]

[ f''(x) = \frac{-4x(x^2+1)^2 - (1 - x^4)(x^2+1)^2 + 4x^3(1 - x^4)}{(x(x^2+1)^2)^2} ]

[ f''(x) = \frac{-4x(x^2+1)^2 - (x^2+1)^2 + x^2(1 - x^4)}{(x(x^2+1)^2)^2} ]

[ f''(x) = \frac{-4x(x^2+1)^2 - (x^4 + 2x^2 + 1) + x^2 - x^6}{(x(x^2+1)^2)^2} ]

[ f''(x) = \frac{-4x^5 - 4x(x^2+1)^2 - x^4 - 2x^2 - 1 + x^2 - x^6}{(x(x^2+1)^2)^2} ]

[ f''(x) = \frac{-5x^6 - 4x^5 - 3x^4 - 6x^2 - 1}{(x(x^2+1)^2)^2} ]

So, the second derivative of ( \ln\left(\frac{x}{x^2+1}\right) ) is:

[ f''(x) = \frac{-5x^6 - 4x^5 - 3x^4 - 6x^2 - 1}{(x(x^2+1)^2)^2} ]

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