What is the derivative of #(x^2 - 4x)/(x-2)^2#?

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Answer 1 · 2016-06-18T05:05:19

#(df)/(dx)=8/(x-2)^3#

To find derivative of #f(x)=(x^2-4x)/(x-2)^2#, we can use quotient rule.

According to quotient rule if #f(x)=(g(x))/(h(x))#

#(df)/(dx)=(h(x)*g'(x)-g(x)*h'(x))/(h(x))^2#

Hence as #f(x)=(x^2-4x)/(x-2)^2#

#(df)/(dx)=((x-2)^2*(2x-4)-(x^2-4x)*2(x-2))/(x-2)^4#

Now we can divide each term by #(x-2)# (as we already have a hole, a discontinuity, at #x=2# and at other places #x-2!=0#) and this results in

#(df)/(dx)=(2(x-2)^2-2(x^2-4x))/(x-2)^3#

= #(2x^2-8x+8-2x^2+8x)/(x-2)^3#

= #8/(x-2)^3#