If #y=x^2e^(1/x)#, what is #(dy)/(dx)#?

Question

Assume #x# is not 0.

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Answer 1 · 2017-07-27T05:50:58

#(dy)/(dx)=e^(1/x)(2x-1)#

Let #u=e^(1/x)#, then #lnu=1/x# and

#1/u(du)/(dx)=-1/x^2# or #(du)/(dx)=-u/x^2=-e^(1/x)/x^2#

Hence foe #y=x^2e^(1/x)#

#(dy)/(dx)=x^2(du)/(dx)+2x xxe^(1/x)#

#=x^2(-e^(1/x)/x^2)+2xe^(1/x)#

#=e^(1/x)(2x-1)#