- #1
Schrodinger's Dog
- 835
- 7
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.
[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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