# What is the second derivative of #f(x) = ln x/x^2 #?

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graph{ln(x)/(x^2) [-0.562, 4.438, -0.77, 1.73]}

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To find the second derivative of ( f(x) = \frac{\ln(x)}{x^2} ), you first need to find the first derivative, and then differentiate it again.

The first derivative of ( f(x) ) can be found using the quotient rule:

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

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

Now, to find the second derivative, differentiate ( f'(x) ) with respect to ( x ):

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

[ f''(x) = \frac{(1 - 2\ln(x) - 2x \cdot \frac{1}{x}) \cdot x^4 - (x - 2x\ln(x)) \cdot 4x^3}{x^8} ]

[ f''(x) = \frac{(1 - 2\ln(x) - 2) \cdot x^4 - (x - 2x\ln(x)) \cdot 4x^3}{x^8} ]

[ f''(x) = \frac{-2\ln(x) \cdot x^4 - 6x^4 + 8x^4\ln(x) + 4x^4\ln(x) - 8x^4}{x^8} ]

[ f''(x) = \frac{-6x^4 - 8x^4 + 2x^4\ln(x)}{x^8} ]

[ f''(x) = \frac{-14x^4}{x^8} + \frac{2\ln(x)}{x^4} ]

[ f''(x) = \frac{-14}{x^4} + \frac{2\ln(x)}{x^4} ]

Therefore, the second derivative of ( f(x) = \frac{\ln(x)}{x^2} ) is ( f''(x) = \frac{-14}{x^4} + \frac{2\ln(x)}{x^4} ).

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