Question #14583

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Answer 1 · 2018-01-09T13:06:51

#y'=dy/dx=1/(2(x-2))-0.5#

#y=0.5ln(x-2)-0.5x-1#

First, differentiate #0.5ln(x-2)#.

#d/dx(0.5ln(x-2))=0.5d/dx(ln(x-2))#

Using the chain rule ,

let #u=x-2#

#0.5d/dx(ln(x-2))=0.5d/dx(ln(u))*d/dx(x-2)#

#0.5d/dx(ln(x-2))=0.5*1/u*1#

#0.5d/dx(ln(x-2))=0.5/u#

Substitute #u=x-2# back in

#0.5d/dx(ln(x-2))=0.5/(x-2)=1/(2(x-2))#

#y'=1/(2(x-2))-(0.5x)'-(1)'#

#y'=1/(2(x-2))-0.5#