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November 29, 2014

Posted by **Francesca** on Sunday, June 3, 2012 at 7:26pm.

y = (-9e^7x) / (5x+3)

Thank you!

- Math -
**Kuai**, Sunday, June 3, 2012 at 7:54pmy = (-9e^7x)/(5x+3)

y' = -9(35x^2+21x+3)*e^7x / (5x+3)^20

- Math -
**MathMate**, Sunday, June 3, 2012 at 7:56pmUse the quotient rule:

d(u/v)=(vdu-udv)/v²

u=(-9e^(7x))

du/dx=-63e^7x

dv/dx=5

so

d(u/v)=((5x+3)(-63e^(7x))+9e^(7x)(5))/(5x+3)^2

=-9(35x+16)e^(7x)/(5x+3)²

- Math -
**Kuai**, Sunday, June 3, 2012 at 8:25pmtypo

There is typo

I do not know where I got power 20

-9( 35x+ 16 ) * e^7x /(5x+3)^2

- Math -
**Francesca**, Sunday, June 3, 2012 at 8:43pmThank you guys!

- Math :) -
**MathMate**, Sunday, June 3, 2012 at 9:00pmYou're welcome!

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