Taking the Second Derivative w/ the Quotient Rule: What if Numerator = 0?

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SUMMARY

The discussion focuses on calculating the second derivative using the quotient rule, specifically for the function s(t) = (t^2 - 2)/(t + 1). The first derivative, v(t), is correctly derived as v(t) = (t^2 + 2t + 2)/(t + 1)^2. The second derivative, a(t), is calculated as a(t) = -2/(t + 1)^3. The key takeaway is that one must differentiate v(t) to find acceleration a(t) rather than substituting directly into v(t).

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  • Practice finding first and second derivatives of various functions
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mathmann
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Just wondering how you take the second derivative when using the quotient rule. After using the quotient rule to get my first derivative, I tried again and the numerator ended up as 0.
 
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Can you show us your exact working?
 
s(t) = t^2 - 2/t + 1, is the object speeing up at 4s?
v(t) = 1.04, a(t) the numerator ended up as a 0. Perhaps I made a calculating error but I went over it a couple times.
 
You mean s(t) = (t^2 - 2)/(t + 1) right?

Then v(t) = (t^2 + 2t + 2)/(t+1)^2

and a(t) = - 2/ (t+1)^3
 
When you evaluate v(t) at some fixed t you get the velocity at that point in time. You are not supposed to differentiate that particular velocity to get the acceleration at that time. You need to work out a(t) first by differentiating the function v(t) before substituting in your fixed t.
 
I understand now.. thanks for the help.
 
No problem :smile:
 

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