anonymous
  • anonymous
dx/dt of ln((t^2)+3)
Mathematics
chestercat
  • chestercat
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myininaya
  • myininaya
\[(\ln(t^2+3))'=\frac{(t^2+3)'}{t^2+3}\]
anonymous
  • anonymous
how about if we dash it again? what would the answer be?
anonymous
  • anonymous
use product rule.

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anonymous
  • anonymous
eerrrr quotient rule i mean
myininaya
  • myininaya
you mean chain rule
anonymous
  • anonymous
could you show me how to do that please?
myininaya
  • myininaya
i showed you above all you need to do is find the derivative of t^2 and derivative of 3 \[\frac{(t^2+3)'}{t^2+3}=\frac{2t+0}{t^2+3}\]
anonymous
  • anonymous
oh no what i mean is if we had to dash it again , like second derivative
myininaya
  • myininaya
dash it again? so you want to find \[(\frac{2t}{t^2+3})'?\]
myininaya
  • myininaya
you will use quotient rule
anonymous
  • anonymous
yes
anonymous
  • anonymous
derivative ln(t2+3) 1/(t^2+3)*(2t)Dt
anonymous
  • anonymous
dash again
anonymous
  • anonymous
|dw:1326853065865:dw|
anonymous
  • anonymous
i want to find second derivative
myininaya
  • myininaya
shiign just put 2nd derivative
myininaya
  • myininaya
all he did was use quotient rule
anonymous
  • anonymous
so the answer is -2t^2 +6 /(t^2+3)^2
anonymous
  • anonymous
is that correct or not?
anonymous
  • anonymous
yes

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