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

Question

anonymous · Answer

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

shadowfiend · Answer

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.

shadowfiend · Answer

Well, chain rule + product rule.

anonymous · Answer

So would t/5t+1 be equal to 2/(5t+1)^2?

shadowfiend · Answer

Heh, I totally messed the thing up there up.

$$\frac{t}{9t + 2} = t(9t + 2)^{-1}$$

shadowfiend · Answer

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

shadowfiend · Answer

You do this by saying:

$$\frac{ds}{dt}t(5t+1)^{-1}$$

$$\begin{align}
(5t + 1)^{-1} + t(-1)(5t + 1)^{-2}(5)\
\frac{1}{5t + 1} - \frac{5t}{(5t + 1)^2}\
\frac{5t + 1}{(5t+1)^2} - \frac{5t}{(5t + 1)^2}\
\frac{5t + 1 - 5t}{(5t + 1)^2}\
\frac{1}{(5t + 1)^2}
\end{align}$$