How do you find the first and second derivative of $ln(x^(1/2))$ ?

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Answer 1 · 2016-11-28T14:05:28

For the first derivative you can use the chain rule:

$(dln(x^(1/2)))/dx= (dln(x^(1/2)))/(d(x^(1/2)))*(dx^(1/2))/dx = 1/x^(1/2)*1/2x^(-1/2)=1/(2x)$

but you can also observe that:

$ln(x^(1/2)) = 1/2lnx$

Either way the second derivative is:

$(d^((2))ln(x^(1/2)))/(d^2x)=(d(1/(2x)))/dx=-1/(2x^2)$

How do you find the first and second derivative of ln(x^(1/2))ln(x^(1/2))?