Simplifying dy/dx for y = e^-2x / x^2 using quotient rule

[Schrodinger's Dog]

dy/dx, where y=\frac{e^{-2x}}{x^2}Using the quotient rule I get

\frac {-2e^{-2x}(x^2) - e^{-2x}(2x)}{x^4}

Simplified I get:-

\frac{e^{-2x}(-2x^2-2x)}{x^4}

The answer is -\frac{2e^{-2x}(x+1)}{x^3}

Simple question can someone run me through the simplification, I'm not quite getting why it's over x^3 here. I'm sure it's just a simple fraction deal, but if someone could break it down nice and simply it would help.

Thanks in advance.

EDIT: sorry I corrected my second step. I accidently added an extra minus sign.

 

[neutrino]

Simple question can someone run me through the simplification, I'm not quite getting why it's over x^3 here. I'm sure it's just a simple fraction deal, but if someone could break it down nice and simply it would help.

You're pulling a 2x out from (2x²+ 2x).

You need to keep track of the minus signs, though.

 

[Schrodinger's Dog]

You're pulling a 2x out from (2x²+ 2x).

You need to keep track of the minus signs, though.

Ah I see if I divide \frac{e^{-2x}(-2x^2-2x)}{x^4} by -2x. I get -1/2x^3 (e^{-2x}(x+1)) which is -\frac {2e^{-2x}(x+1)}{x^3} I'm wondering what my problem was here. Thanks there. Simple really.

 

[neutrino]

Ah I see if I divide \frac{e^{-2x}(-2x^2-2x)}{x^4} by -2x.

Actually, you divide AND multiply by -2x.

 

[Schrodinger's Dog]

Actually, you divide AND multiply by -2x.

Sure that's what I meant, thanks.