anonymous
  • anonymous
trying to find the second derivative in this question......3x-3 over 2x+4. the Vertical Asy is -2 and the Horizontal Asy is 3/2...the xint is (1,0) and the yint is (0,-3/4) please give me step buy step
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
can you find the first derivative? because that is the only hard part
anonymous
  • anonymous
quotient rule give \[\frac{d}{dx}\frac{3x-3}{2x+4}=\frac{(2x+4)\times 3-(3x+3)\times 2}{(2x+4)^2}\]
anonymous
  • anonymous
yes i have the first one all ready

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anonymous
  • anonymous
so you end up with \[\frac{9}{(2x+4)^2}=9(2x+4)^{-2}\] now second should be easy
anonymous
  • anonymous
\[-18(2x+4)^{-3}\times 2=-36(2x+4)^{-3}=-\frac{36}{(2x+4)^3}\]
anonymous
  • anonymous
you can cancel a little because \[(2x+4)^3=(2(x+2))^3=8(x+2)^3\]
anonymous
  • anonymous
im so confused
anonymous
  • anonymous
ok which step? did you get the first derivative?
anonymous
  • anonymous
i got (2x+4)-2(3x-3) over (2x+4)^2
anonymous
  • anonymous
then you have to do the algebra in the numerator
anonymous
  • anonymous
should be \[(2x+4)3-2(3x-3) \]
anonymous
  • anonymous
in front of the 2x+4 there is suppose to be a 3
anonymous
  • anonymous
yes it is the denominator times the derivative of the numerator as the first term
anonymous
  • anonymous
minus the numerator times the derivative of the denominator
anonymous
  • anonymous
if you multiply out the x terms add up to zero and you get 9
anonymous
  • anonymous
so 18 over (2x+4)^2 ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[(2x+4)3-2(3x-3)\] \[6x+12-6x+6=18\] yes
anonymous
  • anonymous
my equation is 3(2x+4)-2(3x-3)
anonymous
  • anonymous
yes multiply out and you will get 18
anonymous
  • anonymous
ok then what do i do to get the second answer
anonymous
  • anonymous
you have \[\frac{18}{(2x+4)^2}\] so rather than using the quotient rule again, rewrite in exponential form as \[18\times (2x+4)^{-2}\] and use the power rule (and the chain rule)
anonymous
  • anonymous
ok so multiple it out?
anonymous
  • anonymous
power rule right? \[\frac{d}{dx}x^n=nx^{n-1}\]
anonymous
  • anonymous
i made a mistake, sorry
anonymous
  • anonymous
\[\frac{d}{dx}18(2x+4)^{-2}=-2\times 18(2x+4)^{-2-1}\times 2\] \[=-72(2x+4)^{-3}=-\frac{72}{(2x+4)^3}\] but you can simplify this
anonymous
  • anonymous
because \[(2x+4)^3=(2(x+2))^3=8(x+2)^3\] and \[\frac{72}{8}=9\]
anonymous
  • anonymous
so your "final answer" will be \[-\frac{9}{(x+2)^3}\]
anonymous
  • anonymous
thank you so much
anonymous
  • anonymous
yw

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