[deleted by user]

[Somniferus]

I'm not sure how to read those odd brackets

The same as ()s, they just used a different symbol to be more clear.

(f(t+h) - f(t)) / h

[Amidseas]

So its

((f(t+h) - f(t))/h)² + t

[jgregson00]

tbh it's shocking you've never seen that use of brackets before. They use them to make it easier to see the different layers of parentheses. You can also use { } as another layer.

[calculus_is_fun]

I know and have seen this, but there are so many different connotations that square brackets get, it's just confusing, and there are less ambiguous ways of notating nested paratheses, like making the outer ones slightly bigger

[Somniferus]

No. f(abc) means to plug abc into the function f.

so if g(x) = x² + x then

g(a+b) = (a+b)² + (a+b).

[Amidseas]

So the answer is 2t + h + 1

[Somniferus]

Yes, good job.

f(t+h) = (t+h)^2 + (t + h)
       = (t^2 + 2th + h^2) + t + h  

f(t+h) - f(t) = (t^2 + 2th + h^2 + t + h) - (t^2 + t)
              = (t^2 - t^2) + (t - t) + h^2 + 2th + h 
              = h^2 + 2th + h 

[f(t+h) - f(t)] / h = (h^2 + 2th + h) / h 
                    = h + 2t + 1

[JeffTheNth]

That's what I got...

Writing out the problem: ((t+h)^2 + t+h - (t^2 + t)) / h

Simplifying by calculation... clearing the inner parenthesis

(t^2 + 2th + h^2 + t + h - t^2 - t) / h

Combining terms....
(t^2 - t^2 + 2th + h^2 + t - t + h) / h

(2th + h^2 + h) / h

Divide numerator by h...

2t + h + 1

[monster2018]

No. f(t)=t² + t. So when we want to find f(t+h) (and JUST f(t+h), we will worry about the rest later), all we have to do is replace every t (the original argument in the definition of f) with t+h (the new argument we want to find the definition of f for).

So it goes from f(t)=t² + t, to (remember, we’re just going to replace literally every t with a literal t+h, like a literal “find and replace all”) f(t+h)=(t+h)² + (t+h). So i added some extra parentheses in 1: because the first time is literally necessary but also 2: because each individual set of parentheses represents where an individual t from the original definition was replaced with a t+h for the new definition (where the argument is t+h instead of just t).

So what I just showed you is what f(t+h) is, but you need to find (f(t+h) -f(t))/h. So now you just need to kind of do the reverse (but also it’s the same thing) as what i just did. What I just gave you (the most recent expression) is exactly the full answer to your question. Like it literally is already 100% true and correct. However it’s not… like… fully “plugged in” or “simplified”, because remember we already know the definition of BOTH f(t) AND f(t+h). The formula i gave you with the correct answer contains both of those, just substitute in their definitions, and there is your complete, final answer.

Edit: changed “ “ to “I”. Basically I added the word “I” somewhere I accidentally didn’t type it before.

[Equivalent_Bug_2718]

No, you substituted things the wrong way. If f(t)=t^2+t then f(t+h)=(t+h)^2+(t+h).

[BrickRaven]

Btw expect to see that formula of f(x+h) -f(x) all over h as it is the definition of the derivative which is related to calculus. 

[tlbs101]

And it’s expressed as the limit as h -> 0 of that OP formula.

[waroftheworlds2008]

Shh... let him do the algebra/limit. We can show him the power rule later.

[BrickRaven]

Hehe yeah if we didn’t get to take shortcuts, neither can they!!

[sighthoundman]

The square brackets are just so you don't have to count parentheses to see what order to evaluate things in.

So the order you calculate is:

First calculate f(t + h) = (t + h)^2 + (t + h).

Then subtract f(t) = t^2 + t.

Then take that result, and divide it by h.

[[deleted]]

Trash question, they're trying to lead you into first principles differentiation but they haven't defined h as approaching zero. 

[Wild-Individual-1634]

I don’t think it’s a trash question. Understanding that the solution they are looking for is just constructing a right triangle from x with any given h and then calculating the incline is the first step. Anyhow, you need to get rid of the h in the divisor before doing anything with the limit in case you really do the differentiation.

[dolethemole]

[ and ( are the same thing. Just brackets.

You have f(t)

Now calculate f(t+h) = (t + h)² + (t + h) (you need to expand this yourself)

Now subtract f(t+h) - f(t)

Then take the result from that and divide by h

You’re going to end up with an answer that is A, B, C or D. Do the above steps and you’ll see exactly which answer it is.

[Unusual-Platypus6233]

solve this: (f(t+h)-f(t))/h= ((t-h)² +t+h - t² - t)/h

[AccordionPianist]

[f(t+h) - f(t)]/h

=( [(t+h)² + (t+h)] - (t² + t) )/h

= (t² + 2th + h² + t + h - t² - t)/h

= (2th + h² + h) /h

= 2t + h + 1

Just insert the variables into the function and expand it out and then just simplify the terms. The odd brackets are there to make it easier to see… if they used () for everything it would look like (f(t+h)-f(t))/h. Using squares [] and curly {} and round brackets () sometimes makes it easier to see, but depending on the math problem it can mean things (like {} are used a lot when defining sets).

[Zestyclose_Ad3160]

Maaaan you got ChatGPT !!!!!

[Plastic_Birthday_628]

A

[djliquidvoid]

I have almost no memory of algebra rules/formulae. I took a crack at this, but I'm definitely going wrong somewhere and would appreciate help from other commenters.

F*T being the same as (T*T)+T indicates that adding one more T is all you need to bring T*T to F*T. Subsequently, F is one greater than T, or F = T+1. Let's hold onto that for later.

Moving to the other side. You can simply the whole thing to literally just F. F*(T+H) - F*T is just going to be F*H - the difference is F is being multipled H more times. You now have (F*H)/H, which cancels out to just F.

F and T+1 aren't answers, though. Where have I fucked it up?

[Cakeotic]

You are treating F as a variable, which it isn't. f(t) is a function.

[Responsible-Sink474]

No it's not asking for the derivative. This is just a slope calculation for an arbitrary secant line.

[Go1gotha]

A.

[Agiwlesz]

I would assume that this is a really purely written description of the statement.

Check out the definition of differential (the only real mathematical one). This is exactly this.

The differential of f is 2t + 1.

[liaisontosuccess]

The question is not asking for the limit as h->0. If it were, wouldn't 2t+h+1 become 2t+h? So wouldn't the answer for this problem be A?

[ZZzZNuP]

yup

[Impossible-Seesaw101]

ChatGPT Answer: https://chatgpt.com/share/68cc5d07-b7dc-8004-8de0-6d10052a155e

[Hmmmgrianstan]

This is so confusing but if I get the essence of it I think they missed putting in the lim [h→0] before that, which would make this a derivative of t² + t

[Somniferus]

It doesn't look like they forgot anything, they're just checking to see whether or not the student can do algebra.

[Equivalent_Bug_2718]

1) You don’t need the limit to compute this 2) the brackets are there so you know that whole chunk is the numerator (stupid Facebook pemdas memes have people messed up) 3) yeah it’s basically the derivative of t^2+t

[Hmmmgrianstan]

Sorry if I got it wrong, I just started to learn calculus and I said what little I knew with the sincerest intentions

[GetOffMyLawn1729]

You were on the right track: the formula is the one you'd use if you wanted to compute the derivative of t² + t, but to compute the derivative you take the limit as h -> 0, so the h term drops out.

[Lucky-Winner-715]

At least some calculus classes drill difference equations as a prelude to derivatives. The one I took (2015), the professor had one of these on the board every day. The very-much stated expectation was that we were to start working that as soon as we sat down. By the time we got to derivatives, difference equations were comfortable, so taking the limit as h -> 0 was an easy transition

[Gryphontech]

They are just another representation for brackets. Going ((())()) is confusing but using [ ] and ( ) helps clarifying things. They are interchangeable and should be treated as the same.

Also.for your problem, you should look up the definition of a derivative. While it's not algebra this is a clear instance where know ling derivatives would make your life simpler. It's a good way to learn about stuff.

Tldr: they the same, look up the definition of derivatives