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In terms of what?

\[\large \frac{d}{dt}\frac{\sqrt t (1+t)}{t^2}\]Does this look accurate? :)

sorry in relation to g(t) and yes that's it

At this point, we haven't taken the derivative yet. But it should be easier to work with from here.

Looks like we'll have to apply the `Product Rule`.
Understand how to do that? :)

Let me rephrase, how did you know to apply the 1/t^2 to the sqrt(t) instead of the (1+t)

Ok. I guess I am just so used to terms having to be applied equally in order to cancel something.

What do we do next with:
\[t ^{-3/2}(1+t)\]

Yeah we have done limits, but I didn't seem to have as much trouble with those for some reason.

So we need to take the derivative of the blue terms.

To take the derivative of the first blue term, we'll simply apply the `Power Rule for Derivatives`.

~The power comes down as a factor in front,
~Decrease the power by 1.

Ok first one looks good :)

Yah since it's inside of a set of brackets, that's probably what's throwing you off.

Have you learned the `Chain Rule` yet? I'm just curious :3
Doesn't affect our problem either way.

hah XD nice

My book says the answer is:
\[-(3/2)t ^{-5/2}-(1/2)t ^{-3/2}\]

Hmm lemme see if I made a boo boo somewhere.

No the problem reads exactly:
\[g(t) = \frac{\sqrt{t}(1+t) }{ t^2 }\]

I think I finally have it... Thank you sooo much!

yay \c:/