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Mathematics
OpenStudy (anonymous):

dx/dt of ln((t^2)+3)

myininaya (myininaya):

\[(\ln(t^2+3))'=\frac{(t^2+3)'}{t^2+3}\]

OpenStudy (anonymous):

how about if we dash it again? what would the answer be?

OpenStudy (anonymous):

use product rule.

OpenStudy (anonymous):

eerrrr quotient rule i mean

myininaya (myininaya):

you mean chain rule

OpenStudy (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 \[\frac{(t^2+3)'}{t^2+3}=\frac{2t+0}{t^2+3}\]

OpenStudy (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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

derivative ln(t2+3) 1/(t^2+3)*(2t)Dt

OpenStudy (anonymous):

dash again

OpenStudy (anonymous):

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