How do I find the derivative of #ln(x-2)/(x-2)#?

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Answer 1 · 2016-01-30T18:02:15

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

Use the quotient rule, which states that

#d/dx[(f(x))/(g(x))]=(f'(x)g(x)-g'(x)f(x))/[g(x)]^2#

Applying this to #ln(x-2)/(x-2)#, we see that\

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

The respective derivatives are:

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

#d/dx[x-2]=1#

Yielding:

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

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