How do you find the second derivative of $ln (X^{\frac{1}{2}})$ $ln(X^(1/2))$ ?

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Answer 1 · 2017-09-08T16:27:37

$- \frac{1}{2} \cdot x^{- 2}, or, - \frac{1}{2 x^{2}} .$ $-1/2*x^(-2), or, -1/(2x^2).$

Let, $f (x) = ln (x^{\frac{1}{2}}) .$ $f(x)=ln(x^(1/2)).$

Using the Rule : $ln (x^{m}) = m ln x,$ $ln(x^m)=mlnx,$ we find,

$f (x) = \frac{1}{2} \cdot ln x .$ $f(x)=1/2*lnx.$

Now, for a constant $k, {k*F(x)}'=k*F'(x),$

$:. f'(x)=1/2{lnx}'=1/2*1/x=1/2*x^(-1).$

Recall that, $f''(x)={f'(x)}'.$

$f''(x)={1/2*x^(-1)}'=1/2{x^(-1)}'.$

But, ${x^n}'=n*x^(n-1),$

$:. f''(x)=1/2*(-1*x^(-1-1))=-1/2*x^(-2), or, -1/(2x^2).$ #

How do you find the second derivative of ln(X^(1/2))ln(X12)ln(X^(1/2))?