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anonymous

  • 5 years ago

Why is ds/dt of s=t/9t+2 = 2/(9t+2)^2 and not the derivative 9x^2+2?

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  1. anonymous
    • 5 years ago
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    Basically, I answered the question with 9x^2+2, but I was wrong and I'm trying to figure out why. :(

  2. shadowfiend
    • 5 years ago
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    Well, when you have an equation divided by another equation, you have to apply the quotient rule or the chain rule. I prefer the chain rule, personally. So: \[\frac{t}{9t + 2} = t^{-1}(9t + 2)\] If you apply the chain rule to that, the answer might make a bit more sense.

  3. shadowfiend
    • 5 years ago
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    Well, chain rule + product rule.

  4. anonymous
    • 5 years ago
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    So would t/5t+1 be equal to 2/(5t+1)^2?

  5. shadowfiend
    • 5 years ago
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    Heh, I totally messed the thing up there up. \[\frac{t}{9t + 2} = t(9t + 2)^{-1}\]

  6. shadowfiend
    • 5 years ago
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    No, it wouldn't derive to that. It would derive to 1/(5t + 1)^2.

  7. shadowfiend
    • 5 years ago
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    You do this by saying: \[\frac{ds}{dt}t(5t+1)^{-1}\] \[\begin{align} (5t + 1)^{-1} + t(-1)(5t + 1)^{-2}(5)\\ \frac{1}{5t + 1} - \frac{5t}{(5t + 1)^2}\\ \frac{5t + 1}{(5t+1)^2} - \frac{5t}{(5t + 1)^2}\\ \frac{5t + 1 - 5t}{(5t + 1)^2}\\ \frac{1}{(5t + 1)^2} \end{align}\]

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