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f(x)=xln(x)

this function is treated as

u*v

u=x

v=ln(x)

The derivative of a product is

f'(x)=u*v'+v*u'

plugging the values back in you get:

f'(x)=(x*dlnx/x)+(ln*dx/dx)

The derivative of lnx=1/x

x=u

dlnu/dx=(1/u)*(du/dx)

dx/dx=1

x=u

dun/dx=nun-1

dx1/dx=1*x1-1

= x0=1

f'(x)=x*(1/x)+lnx*1

f'(x)=1+lnx

Now for the second derivative

f''(x)=d1/dx+dlnx/dx

the derivative of a constant, such as 1, is 0

and knowing that the derivative of lnx=1/x you get

f''(x)=(1/x)

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Q: What is the second derivative of x ln x?
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