anonymous
  • anonymous
Find the second derivative of the function g(t) = -(4/((t+2)^2))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
-4(t+2)^-2
amistre64
  • amistre64
g' = 8(t+2) g'' = 8
amistre64
  • amistre64
ugh...soo close

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More answers

anonymous
  • anonymous
thats a fraction
amistre64
  • amistre64
g' = 8(t+2)^-3 g'' = -24(t+2)^-4
anonymous
  • anonymous
hi amistre! -4/ (t+2)^2
amistre64
  • amistre64
\[g''=\frac{-24}{(t+2)^4}\]
anonymous
  • anonymous
oh nvm i see what u did sorry
amistre64
  • amistre64
product rule tends to be easier to follow that quotient rule
anonymous
  • anonymous
oh i used the quotient rule
anonymous
  • anonymous
for g't i got 8t+16/(t^2+4t+4)^2
amistre64
  • amistre64
we could test it by taking the -4 aside: \[{1\over(t+2)^2}\iff \frac{(t+2)^2 0 - 2(t+2)}{(t+2)^4} \iff \frac{-2}{(t+2)^3}\]
amistre64
  • amistre64
but it just follows the product rule of it lol
anonymous
  • anonymous
for g''(t) i got -16t^4-192t^3-480t^2-384t-1008/ (t^4+8t^3+12t^2+32t+16)
amistre64
  • amistre64
\[g' = {8 \over (t+2)^3} \iff 8(t+2)^{-3} \] right?
anonymous
  • anonymous
hmm confused
anonymous
  • anonymous
so what your saying is that our answers are the same? is my answer correct?
amistre64
  • amistre64
i dont see your answer as being correct; it looks like you got lost in the details
amistre64
  • amistre64
\[\frac{8}{(t+2)^3}\] doesnt turn into that
anonymous
  • anonymous
i'll just use the product rule
amistre64
  • amistre64
\[g'' = {-8.3(t+2)^2 \over (t+2)^6} \iff {-24 \over (t+2)^4}\]
anonymous
  • anonymous
because it can get pretty contangeled like this
amistre64
  • amistre64
product rule is simplest :)
anonymous
  • anonymous
thank you so much :)
amistre64
  • amistre64
yw :)

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