if y=x(ln x)^2, then y'=?

Question

if y=x(ln x)^2, then y'=

Answer 1

Using the product rule
f[x]= x
g[x]= (lnx)^2
f'[x]= 1
g'[x] = 2lnx*1/x

Plugging these values into the product rule equation of
f'[x]g[x] + f[x]g'[x], we get
f'[x] = (lnx)^2 + x(2lnx *1/x)
f'[x] = (lnx)^2 +2lnx

Answer 2

Use the product rule: y=xz so y'=xz'+x'z
y=x(ln(x))^2 Use product rule
y'=x((2ln(x))/x) + (1)(ln(x))^2 Simplify

y'=2ln(x) + (ln(x))^2

Answer 3

if y=x(ln x)^2, then y'=
(ln(x)^2+2*ln(x))*x/x=
(x*ln(x)^2+2*x*ln(x))/x

Answer 4

y ' = (ln(x))^2 + 2ln(x)