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garner123
Calculus: Find the derivative of: g(t) = (sqrt(t)(1+t))/t^2 I am familiar with derivatives, but with this one I didn't know where to start.
\[\large \frac{d}{dt}\frac{\sqrt t (1+t)}{t^2}\]Does this look accurate? :)
sorry in relation to g(t) and yes that's it
Let's simplify it down before taking the derivative. Rewrite the square root as a rational expression. \[\large \frac{d}{dt}\frac{t^{1/2}(1+t)}{t^2}\]Then let's divide the t's. Remember how to divide terms that have the same base?? We subtract the exponents.\[\large \frac{d}{dt}t^{(1/2-2)}(1+t)\]Which simplifies our expression to,\[\large \frac{d}{dt}t^{-3/2}(1+t)\]
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? :)
Oh right... That's one part I was doing wrong... Vaguely, but I have to ask why (1+t) hasn't been changed by the denominator when it was brought up...
Let me rephrase, how did you know to apply the 1/t^2 to the sqrt(t) instead of the (1+t)
\[\large \frac{a\cdot c}{b} \qquad = \qquad \frac{a}{b}\cdot c \qquad = \qquad a\cdot \frac{c}{b}\] These are all equivalent. We could have applied it to the other term, but it wouldn't have helped us. I was about to say "because division is commutative" but I don't think that's accurate. Blah ignore that sentence XD
Ok. I guess I am just so used to terms having to be applied equally in order to cancel something.
Yah I really dislike the way they teach lower levels of math. PEMDAS or whatever it's called. Because there are so many little manipulations you can do to bend the rules :\ You'll get pretty comfortable doing them as you go through calculus. Have you done work with limits yet? :) Those require quite a few little math tricks.
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.
We'll apply the `Product Rule`.\[\large (uv)'=u'v+uv'\]The terms with primes of them are the ones we have to take a derivative of. \[\large \color{royalblue}{\frac{d}{dt}}t^{-3/2}(1+t) \qquad =\] \[\large \color{royalblue}{\frac{d}{dt}t^{-3/2}}(1+t)\qquad + \qquad t^{-3/2}\color{royalblue}{\frac{d}{dt}(1+t)}\]
I used two different notations, hopefully that's not confusing. It's good to get comfortable with both :)
So we need to take the derivative of the blue terms.
While you're learning these, it might be a good idea to `set up` the Product Rule as I have done here, just to keep things organized. It's a bit of an intermediate step, but whatev :D
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.
I've got \[-(3/2)t ^{-5/2}\] for the first part. I am lost on taking the derivative of (1+t) which I know is probably easier.
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.
I think we briefly went over it today but I am not sure because my prof. doesn't refer to rules by their proper names. He usually explains it in another way as a way of 'simplifying' it. I love his class, but it can be easy to get lost.
We'll just take the derivative of the contents inside of the brackets. If something was being applied to the outside of the brackets, like a square, we would have to deal with that also. But we're ok to jump right inside in this case. We'll just take the derivative of each term individually, and keep the addition between them. The `1` is a constant value. Do you remember the derivative of a constant? It's easy to forget I suppose. :) Here is how I like to think about it: A constant is a value that does not change. When you take the derivative, you're asking, "As t changes, how much does `a value that doesn't change` change?"
Maybe that's weird though -_- If you want, you can use the power rule to deal with constants. at least to get a handle on what's going on. \[\large 1 \quad = t^0\]When we take the derivative, a 0 comes down, so it doesn't matter what the new power is, because it's being multiplied by 0.
\[\large \frac{d}{dt}\left(1+t\right) \qquad \rightarrow \qquad \frac{d}{dt}\left(t^0+t^1\right)\] Yah maybe this is a way to get a feel for what's going on :O Apply the power rule from here.
\[\large \frac{d}{dt}\left(t^0+t^1\right) \qquad \rightarrow \qquad \left(0t^{-1}+1t^0\right) \qquad \rightarrow \qquad \left(0+1\right)\]
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.
Hmm no that's way wrong. I entered it into Wolfram to check. Did we make a mistake at the start maybe? Was the entire top suppose to be under the square root?
No the problem reads exactly: \[g(t) = \frac{\sqrt{t}(1+t) }{ t^2 }\]
\[\large -\frac{3}{2}t^{-5/2}(t+1)+t^{-3/2} \qquad \rightarrow \qquad \frac{1}{t^{3/2}}-\frac{3(t+1)}{2t^{5/2}}\]We need a common denominator. We need to multiply the first term by 2t/2t.\[\large \frac{1}{t^{3/2}}-\frac{3(t+1)}{2t^{5/2}} \qquad \rightarrow \qquad \frac{2t}{2t^{5/2}}-\frac{3(t+1)}{2t^{5/2}}\] Which becomes,\[\large \frac{2t-3t+3}{2t^{5/2}} \qquad \rightarrow \qquad \frac{3-t}{2t^{5/2}}\]Which is equal to,\[\large \frac{3}{2}t^{-5/2}-\frac{1}{2}t^{-3/2}\]
Yep yer right, it is equal to that, sorry bout that. I forgot that when you're dealing with a book answer, they will often WAAAAAAAY oversimplify things, to the point where you can't even recognize the answer.
I didn't distribute the negative to the 3, woops* the first term should have a negative on it as well :) as you posted.
I find that rather irritating because your teacher will not want you to simplify the answer down that much. He'll just want to see that you know how to do the differentiation. It makes it really hard to check your work.
\[(1+t) + t^{-3/2}\] Uh-oh.. Where did that come from? I'm sorryy and thank you for everything thus far.
\[\large \color{royalblue}{\frac{d}{dt}t^{-3/2}}(1+t)\qquad + \qquad t^{-3/2}\color{royalblue}{\frac{d}{dt}(1+t)}\] Taking the derivative of the first blue term gives us,\[\large \color{royalblue}{-\frac{3}{2}t^{-5/2}}(1+t)\qquad + \qquad t^{-3/2}\color{royalblue}{\frac{d}{dt}(1+t)}\] Taking the derivative of the other blue term gives us,\[\large \color{royalblue}{-\frac{3}{2}t^{-5/2}}(1+t)\qquad + \qquad t^{-3/2}\color{royalblue}{(1)}\] Which gave us,\[\large -\frac{3}{2}t^{-5/2}(t+1)\qquad +\qquad t^{-3/2}\]
You should leave your answer like that, there's no reason to go any further. Unless your teacher asks you to of course :)
I think I finally have it... Thank you sooo much!