Question

How do you find the second derivative of `f(x)=x^2 lnx` ?

Answer 1

`f''(x)=2ln(x)+3`

Explanation:

By the product rule we get
`f'(x)=2xln(x)+x^2*17x`
simplifying
`f'(x)=2xln(x)+x`
`f''(x)=2ln(x)+2x*1/x+1`
simplifying we get
`f''(x)=2ln(x)+3`

Answer 2

`f''(x)=3+2lnx`

Explanation:

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

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

`f'(x)=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"`

`g(x)=x^2rArrg'(x)=2x`

`h(x)=lnxrArrh'(x)=1/x`

`f'(x)=x^2. 1/x+2xlnx=x+2xlnx`

`"differentiate "2xlnx" using the "color(blue)"product rule"`

`g(x)=2xrArrg'(x)=2`

`h(x)=lnxrArrh'(x)=1/x`

`d/dx(2xlnx)=2x . 1/x+2lnx=2+2lnx`

`f''(x)=1+2+2lnx=3+2lnx`