Question

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

Answer 1

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

Explanation:

`f(x) = xln^2x`

`f'(x) = x*2lnx*1/x + ln^2x*1` [Product rule, chain rule and power rule]

`= 2lnx+ln^2x`

`= lnx(lnx+2)`

`f''(x) = lnx*d/dx(lnx+2) +d/dx lnx* (lnx+2)` [Product rule]

`= lnx(1/x+0) + 1/x*(lnx+2)` [Standard differential]

`= 1/x*(2lnx+2)`

`= 2/x(lnx+1)`

Answer 2

Use the product rule to get: `y'=2ln(x)+ln(x)^2`
and the same for getting `y''=2/x(ln(x)+1)`

Explanation:

Step 1. Break the original into two functions

`u=x` and `v=ln(x)^2`

Step 2. Differentiate each `u` and `v`

`d/dx(u)=d/dx(x)=1`

`d/dx(v)=d/dx(ln(x)^2)=2*ln(x)*1/x`

Step 3. Use the Product Rule to solve

Product Rule: `dy/dx=u (dv)/dx+v (du)/dx`

Plugging in: `dy/dx=x*2ln(x)1/x+ln(x)^2*1`
`dy/dx=2ln(x)+ln(x)^2=ln(x)(2+ln(x))`

Step 4. Differentiate the answer from Step 3.
`(d^2y)/dx^2=ln(x)1/x+(2+ln(x))1/x`
`=ln(x)/x+2/x+ln(x)/x`
`=1/x(2ln(x)+2)`
`=2/x(ln(x)+1)`