Question

How do you find the first and second derivative of `(lnx)^2`?

Answer 1

`"see explanation"`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x)xxg'(x)larrcolor(blue)"chain rule"`

`rArrd/dx((lnx)^2)`

`=2lnx xxd/dx(lnx)=(2lnx)/xlarrcolor(blue)"first derivative"`

`"differentiate first derivative to obtain second derivative"`

`d/dx((2lnx)/x)`

`"differentiate using the "color(blue)"product rule"`

`"given "y=g(x)h(x)" then"`

`dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"`

`"express "(2lnx)/x=2lnx .x^-1`

`g(x)=2lnxrArrg'(x)=2/x`

`h(x)=x^-1rArrh'(x)=-x^-2=-1/x^2`

`rArrd/dx(2lnx .x^-1)`

`=-2lnx . 1/x^2+1/x . 2/x`

`=2/x^2-(2lnx)/x^2larrcolor(blue)"second derivative"`