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dx/dt of ln((t^2)+3)
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\[(\ln(t^2+3))'=\frac{(t^2+3)'}{t^2+3}\]
how about if we dash it again? what would the answer be?
use product rule.
eerrrr quotient rule i mean
you mean chain rule
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could you show me how to do that please?
i showed you above all you need to do is find the derivative of t^2 and derivative of 3 \[\frac{(t^2+3)'}{t^2+3}=\frac{2t+0}{t^2+3}\]
oh no what i mean is if we had to dash it again , like second derivative
dash it again? so you want to find \[(\frac{2t}{t^2+3})'?\]
you will use quotient rule
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yes
derivative ln(t2+3) 1/(t^2+3)*(2t)Dt
dash again
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