What are the first and second derivatives of $f(x)=lnx/x^2$ ?

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Answer 1 · 2015-12-22T21:21:33

Using quotient rule, which states that, for a function $y=f(x)/g(x)$ , $(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2$

$(dy)/(dx)=((1/cancel(x))x^cancel(2)-lnx(2x))/(x^4)=(cancel(x)(1+2lnx))/x^(cancel(4)3)$

$(dy^2)/(d^2x)=((2/x)x^3-(1+2lnx)*3x^2)/x^6$

$(dy^2)/(d^2x)=(x^2(2-3(1+2lnx)))/x^6$

$(dy^2)/(d^2x)=(x^2(2-3-6lnx))/x^6$

$(dy^2)/(d^2x)=(-1-6lnx)/x^4$

$(dy^2)/(d^2x)=-(1+6lnx)/x^4$

What are the first and second derivatives of f(x)=lnx/x^2f(x)=lnx/x^2?