Answer:
[tex]\frac{df(x)}{dx}=2(1+lnx)[/tex]
Step-by-step explanation:
From our knowledge of derivatives we use the product rule approach, the product rule is stated below
For f(x)=u(x)v(x) the derivative is express as
[tex]\frac{df(x)}{dx}=u(x)\frac{dv(x)}{dx}+v(x)\frac{du(x)}{dx}\\[/tex]
hence from the question f(x)=2xlnx,
we can assign u(x)=2x and v(x)=lnx
also the derivative of u(x)=2x is simply
[tex]\frac{du(x)}{dx}=2[/tex]
and the derivative of v(x)=lnx can be express using the general log derivative i.e
[tex]y=lnx\\\frac{dy}{dx}=\frac{1}{x} \\[/tex]
hence we have [tex]\frac{dv(x)}{dx}=\frac{1}{x} \\[/tex]
if we substitute value into the general product expression stated earlier we arrive at
[tex]\frac{df(x)}{dx}=2x\frac{1}{x}+lnx*2\\\frac{df(x)}{dx}=2+2lnx\\\frac{df(x)}{dx}=2(1+lnx)[/tex]
