How do you find the second derivative of $f (x) = ln x + \sqrt{162}$ $f(x)=lnx+sqrt(162)$ ?

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Answer 1 · 2015-03-01T15:11:54

The answer is: $y''=-1/x^2$ .

Since $sqrt162$ is a number, so:

$y'=1/x=x^-1$

and

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

How do you find the second derivative of f(x)=lnx+sqrt(162)f(x)=lnx+√162f(x)=lnx+sqrt(162)?