The derivative of the derivative of xlnx; y''=?

Question

My answer was 1/x. Just checking to see if it's right or not.

Answer 1

d/dx(xlnx) = d/dx(x)*lnx + x*d/dx(lnx) = lnx + x*(1/x) = lnx + 1
Then, d^2/dx^2(xlnx) = d/dx(lnx + 1) = d/dx(lnx) + d/dx(1) = 1/x + 0 = 1/x

So you are correct.

Answer 2

you need to use the product rule for derivatives, namely f(x)g'(x) + g(x)f'(x). In this problem, choose x be represented by f(x) and ln(x) to be represented by g(x). Now, if you took the derivative of f(x) (which is x here), you would get 1. Do the same for g(x) = (ln(x)) which is 1/x. Now you have everything you need for the product rule's formula, fg' + gf'. So the derivative is x*(1/x) + ln(x)*(1), which can be reduced to 1 + ln(x).

Answer 3

Hi Dear Burkehud

{ If f(x) = lnx, then f’(x) = 1/x
if f(x) =u(x)*v(x)
so f '(x)= u(x) * v'(x) + u '(x)*v(x) }...

Part 1
f(x) = x ln(x)
& u(x) = x , u '(x) = 1
v(x) = lnx , v '(x) = 1/x
f '(x) =( x * 1/x ) + (1 * lnx) = 1 + lnx

Part 2
{ if a is a real number , f '(a) = 0 } ...
f '(x) = 1 + lnx
f "(x) = 0 + 1/x
f "(x) = 1/x

Good Luck dear.

Answer 4

y=xlnx
y'=x1/x+lnx =1+lnx
y"=1/x

Answer 5

y=xlnx
y'=x*1/x+lnx*1
=1+lnx
y"=1/x

Answer 6

y=x ln x
y'=ln x + 1
y''=1/x

Yes, you are correct.

Answer 7

yes you are right.