How do you find the first and second derivative of $ln (\frac{x}{20})$ $ln(x/20)$ ?

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Answer 1 · 2017-01-18T09:54:46

$\frac{d}{d x} ln (\frac{x}{20}) = \frac{1}{x}$ $d/(dx) ln(x/20) = 1/x$

Based on the properties of logarithms:

$ln (\frac{x}{20}) = ln x - ln 20$ $ln(x/20) = lnx -ln20$

so that:

$\frac{d}{d x} ln (\frac{x}{20}) = \frac{d}{d x} ln (x) = \frac{1}{x}$ $d/(dx) ln(x/20) = d/(dx) ln(x) = 1/x$

Answer 2 · 2017-01-18T09:56:59

We use the Rule $ln (\frac{a}{b}) = ln a - ln b,$ $ln(a/b)=lna-lnb,$ and get, $y = ln x - ln 20 .$ $y=lnx-ln20.$

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

$=1/x=x^-1.$

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

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

How do you find the first and second derivative of ln(x/20)ln(x20)ln(x/20)?