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

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

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

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

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