Question

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

Answer 1

We'll use quotient rule and product rule

Explanation:

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`