#x^2 ln(1/x)#?

Question

find derivative.

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Answer 1 · 2018-04-08T17:53:39

#(d(x^2 ln(1/x)))/dx = -2xln(x)-x#

Use the property of logarithms #ln(1/A) = -ln(A)#

#(d(x^2 ln(1/x)))/dx = -(d(x^2 ln(x)))/dx#

#(d(uv))/dx = (du)/dxv+u (dv)/dx#

where #u = x^2# and #v = ln(x)# then #(du)/dx = 2x# and #(dv)/dx = 1/x#:

#(d(x^2 ln(1/x)))/dx = -(2xln(x)+x^2 1/x)#

Simplify:

#(d(x^2 ln(1/x)))/dx = -2xln(x)-x#