What is the derivative and the second derivatie of x^2lnx? and what are the critical points of f' and f''?

Question

anonymous · Answer

I know to use product rule. I just need to confirm my answer.

anonymous · Answer

I'm looking at it now, but it's not the product rule, as you don't have two distinct expressions MULTIPLED together, one expression (x), is raised to the POWER of another expression (2lnx).

mimi_x3 · Answer

$$x^2lnx$$
$$\large 2xlnx + \frac{x^{x}}{x}   = 2xln +x$$
I think that its the product rule..

anonymous · Answer

Oh I"m sorry, I should have noted it better. It's (x^2)(lnx). My bad

anonymous · Answer

How would I find critical points? I know you have to make y=0

anonymous · Answer

and what is the second derivative of that?

anonymous · Answer

ok, yes, I misread your question.
y = (x^2).(ln x)
y' = (x^2).(1/x) + (2x).(ln x) = x + 2x.lnx
y'' = 1 + 2x.(1/x) + 2lnx = 3 + 2lnx
To find critical points:
y = 0 gives x-intercept
y' = 0 gives stationary points (turning points & horizontal inflextions)
use the 2nd derivative to determine the nature of the above
y''=0 gives points of inflexion

anonymous · Answer

So the only critical point for f'(x) is 0?

anonymous · Answer

f'(x) = x+2xlnx = x(1+2lnx) = 0 when x = 0, and when 1+2lnx = 0
1+2lnx = 0
2lnx = -1
lnx = -1/2
x = e^(-1/2)
be thorough!

anonymous · Answer

So the two critical points are 0 and e^0.5?

anonymous · Answer

0 and e to the power MINUS 0.5

anonymous · Answer

Sorry I forgot that minus :P
so for the second derivative, the critical point would be e^-3/2?

anonymous · Answer

what you refer to as "critical points" might be different to me (I'm in Australia), & we refer to intercepts, stationary points, & inflexions - all of which, personally, I would regard as "critical" to getting a good sketch.
2nd derivative: yes, you're right there.

anonymous · Answer

Critical Point: A critical point on a graph occurs at x=c if f(c) is defined, and if and only if f'(c) either is zero or is undefined

anonymous · Answer

so for the second derivative, we would set f''(x) equal to zero
so 0=3+2lnx

and I got x to be e^-3/2. is that right?

anonymous · Answer

and is that the ONLY critical point?