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

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Answer 1 · 2017-09-08T16:27:37

# -1/2*x^(-2), or, -1/(2x^2).#

Let, #f(x)=ln(x^(1/2)).#

Using the Rule : #ln(x^m)=mlnx,# we find,

# 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).##