- 668

- 6

dy/dx, where [tex]y=\frac{e^{-2x}}{x^2}[/tex]

Using the quotient rule I get

[tex]\frac {-2e^{-2x}(x^2) - e^{-2x}(2x)}{x^4}[/tex]

Simplified I get:-

[tex]\frac{e^{-2x}(-2x^2-2x)}{x^4}[/tex]

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

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.

Using the quotient rule I get

[tex]\frac {-2e^{-2x}(x^2) - e^{-2x}(2x)}{x^4}[/tex]

Simplified I get:-

[tex]\frac{e^{-2x}(-2x^2-2x)}{x^4}[/tex]

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

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.

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