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What are the first and second derivatives of # g(x) =(lnx)^2-ln(x^2)#?

Answer 1

Cameron Alexander

#d/dxg(x) = (2(ln(x)-1))/x#

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

We will use

The chain rule: #d/dxf(g(x)) = f'(g(x)g'(x)#

The quotient rule: #d/dxf(x)/g(x) = (f'(x)g(x) - f(x)g'(x))/g^2(x)#

#d/dx (f(x) +- g(x)) = f'(x) +- g'(x)#

#d/dxln(x) = 1/x#

#d/dx x^n = nx^(n-1)#

#g'(x) = d/dx(ln^2(x) - ln(x^2)) = d/dx(ln^2(x)) - d/dx(ln(x^2))#

#=2ln(x)(d/dxln(x)) - 1/(x^2)(d/dxx^2)# (by the chain rule)

#= 2ln(x)(1/x) - 1/x^2(2x)#

#= (2ln(x))/x - 2/x#

#= (2(ln(x) - 1))/x#

To find the second derivative, we take the derivative of #g'(x)# using the quotient rule.

#g''(x) = d/dx(2(ln(x)-1))/x#

#= ((d/dx2(ln(x)-1))x - 2(ln(x)-1)(d/dxx))/x^2#

#=((2(1/x))x - 2(ln(x)-1)(1))/x^2#

#= (2 - 2ln(x) + 2)/x^2#

#= (4-2ln(x))/x^2#

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Answer 2

Aurora Alvarado

The first derivative of ( g(x) = (\ln(x))^2 - \ln(x^2) ) is:

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

The second derivative of ( g(x) = (\ln(x))^2 - \ln(x^2) ) is:

[ g''(x) = 2 \cdot \left( \frac{1}{x} \right)^2 - 2 \cdot (-1) \cdot \frac{1}{x^2} ]

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