Question

Find the second derivative y= x^2lnx

Answer 1

Here is y'
\[y'= 2 x \ln x + x^2 \frac 1 x= 2 x \ln x +x
\]
Try to do y''

Answer 2

I use the product rule in my previous post.

Answer 3

The second derivative of y=x^2lnx is 2lnx+3.

Answer 4

Ive gotten to 2ln + (1/x+1)*2x
i dont see how that simplifies to 3
could someone explain?

Answer 5

\[y''=(2x \ln(x)+x)'=(2x \ln(x) )' +(x)'=2(x \ln(x))'+1=2(1 \ln(x)+x \frac{1}{x})+1\]

Answer 6

I used the following rules:
(f+g)'=f'+g'
(fg)'=f'g+fg'
(cf)'=cf' or we can even say (cfg)'=c(fg)'=c(f'g+fg')