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How do you find the first and second derivative of # ln(ln x)#?

Answer 1

Ezekiel Alexander

#1/(xlnx)#

First derviative: #d/dxln(ln(x))#

Chain rule: #d/dxf(g(x))=f'(g(x))*g'(x)#

Let g=ln(x)

#f(g)=ln(g), g(x)=ln(x)#

#f'(g)=1/g, g'(x)=1/x#

#f'(x)=1/ln(a)#

#f'(g(x))*g'(x)=(1/ln(x)*1/x)#

#1/(xlnx)#

Second Derivative: #d/dx(1/ln(x)*1/x)#

Product rule: #f(x)g'(x)+f'(x)g(x)#

#f(x)=1/lnx, g(x)=1/x#

#1/lnxd/dx(x^-1)+d/dx(lnx)^-1(1/x)#

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

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Answer 2

Adam Adkins

To find the first derivative of ln(ln(x)), use the chain rule:

d/dx [ln(ln(x))] = (1/ln(x)) * (1/x)

To find the second derivative, apply the chain rule again:

d^2/dx^2 [ln(ln(x))] = (1/ln(x)) * (-1/x^2) + (1/ln(x)^2) * (1/x)

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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.