How do you find the first and second derivative of #ln(x/20)#?

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Answer 1 · 2017-01-18T09:54:46

#d/(dx) ln(x/20) = 1/x#

Based on the properties of logarithms:

#ln(x/20) = lnx -ln20#

so that:

#d/(dx) ln(x/20) = d/(dx) ln(x) = 1/x#

Answer 2 · 2017-01-18T09:56:59

We use the Rule #ln(a/b)=lna-lnb,# and get, #y=lnx-ln20.#

#:." The First Derivative "y'=(lnx)'-(ln20)'=1/x-0#

#=1/x=x^-1.#

#"Next, since, "(x^n)'=nx^(n-1)," the Second Derivative "y''#

#=(y')'=(x^-1)'=-1x^(-1-1)=-1/x^2#.