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

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Question

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

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Evelyn Agard · Accepted Answer

#(6ln(x)-5)/(x^4)#

By the quotient rule, the first derivative is #f'(x)=(x^2 * 1/x - ln(x) * 2x)/(x^4)=(x-2xln(x))/(x^4)=(1-2ln(x))/(x^3)#. Use the quotient rule again to find the second derivative:

#f''(x)=(x^3 * -2/x - (1-2ln(x)) * 3x^2)/(x^6)#

#=(-2x^2-3x^2+6x^2ln(x))/(x^6)=(6ln(x)-5)/(x^4)#

For extra interest, the fact that #f'(x)=(1-2ln(x))/(x^3)# implies that #f# is increasing for #0 < x < e^{1/2} approx 1.65# and decreasing for #x > e^(1/2)#, with a local maximum value of #f(e^{1/2})=(1/2)/((e^{1/2})^2)=1/(2e) approx 0.18# at #x=e^{1/2} approx 1.65#. (Make sure you check all this!)

The fact that #f''(x)=(6ln(x)-5)/(x^4)# implies that #f# is concave down for #0 < x < e^{5/6} approx 2.30# and concave up for #x > e^{5/6}#, with an inflection point at #(x,y)=(e^{5/6},f(e^{5/6}))=(e^{5/6},(5/6)/((e^{5/6})^2))=(e^{5/6},5/(6e^{5/3})) approx (2.30,0.16)#. (Make sure you check all this!)

Here's the graph to allow you to see these features.

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

Charles Arocha · Answer

undefined 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} $.