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:

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

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

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