0
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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How do you do exponential functions?
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