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[Spehss _]

Since f'(x^2) = 2x, not 1x, it would seem that graphing is in fact necessary for you to grasp whatever it wants you to.

You got 2x from the equations to calculate f'(x), I'm assuming.

Guess I misunderstood what the graphs of f'(x) represent. Or something. Thought I had it figured out so the equations weren't necessary.

[Reelya]

you can calculate the point tangent by looking at two close-together points and working out the line between them. e.g. if you want the rate of change at x=1, work out the slope of the line from x= 0.99 to x = 1.01, you get the idea. Then you make a table of the slopes at multiple points, e.g. x=1, x=2, x=3,x=4, and you infer the relationship.

So, for example, you work out x^2 at 0.99 and 1.01, then work out how fast that increases in relation to x, i.e. (1.01^2-0.99^2)/0.02, there's your point tangent (aka the derivative). Then repeat that process for points close to 2.0, 3.0, 4.0 etc. The derivative at a point is the limit of those little lines (dy/dx) as the length of dx approaches zero.

[Spehss _]

My calculus textbook says that both lim(x->a)((f(x)-f(a))/(x-a) = f'(x) and lim(h->0)((f(x+h)-f(x))/h) = f'(x)

So I could use either lim(x->a)((f(x)-f(a))/(x-a) or lim(h->0)((f(x+h)-f(x))/h) to calculate f'(x) and I should get the same result, right?

[frostshotgg]

Yes. One of them is a rewritten version of the other. Generally speaking the h->0 is easier to use, because you should have all the h's in the top factor out and cancel with the bottom one so there's no 0s.

[Arx]

Generally speaking the h->0 is easier to use, because you should have all the h's in the top factor out and cancel with the bottom one so there's no 0s.

Eeh. Not really. It happens sometimes, though, I'll grant you.

[Spehss _]

I was asking since solving f(x)=x^4 for f'(x) while using the h>0 method involves factoring (or whatever the term is) (x+h)^4 which is a pain in the butt and x->a seemed easier.

Went ahead and used both methods though and got two separate answers (x^3-a^3 and 4x^3). Pretty sure I did the x->a method wrong so I guess I'll stick with h->0.

[MagmaMcFry]

I was asking since solving f(x)=x^4 for f'(x) while using the h>0 method involves factoring (or whatever the term is) (x+h)^4 which is a pain in the butt and x->a seemed easier.

Went ahead and used both methods though and got two separate answers (x^3-a^3 and 4x^3). Pretty sure I did the x->a method wrong so I guess I'll stick with h->0.

Yeah, you canceled out things you shouldn't have: (x⁴-a⁴)/(x-a) is NOT x³-a³, but actually x³+ax²+a²x+a³. Also you forgot to take the limit: The limit of x³+ax²+a²x+a³ as a approaches x is obviously x³+xx²+x²x+x³ = 4x³.

[Arx]

I don't suggest using this as an answer until you officially 'learn' it, but in general the derivative of a term axⁿ will be anx^n-1. It makes it quick and easy to check answers you get from first principles.

[frostshotgg]

I was asking since solving f(x)=x^4 for f'(x) while using the h>0 method involves factoring (or whatever the term is) (x+h)^4 which is a pain in the butt and x->a seemed easier.

Went ahead and used both methods though and got two separate answers (x^3-a^3 and 4x^3). Pretty sure I did the x->a method wrong so I guess I'll stick with h->0.

You won't be using the limit definition for very long. The point is to demonstrate where it comes from. Pretty soon you'll learn rules like the one Arx mentioned.

[Reelya]

My calculus textbook says that both lim(x->a)((f(x)-f(a))/(x-a) = f'(x) and lim(h->0)((f(x+h)-f(x))/h) = f'(x)

So I could use either lim(x->a)((f(x)-f(a))/(x-a) or lim(h->0)((f(x+h)-f(x))/h) to calculate f'(x) and I should get the same result, right?

They're exactly the same thing. Just set a=x-h and one turns into the other.

~~~

Yeah, and factoring polynomials doesn't work like that.

(x^4-a^4)/(x-a) does not equal x^3 - a^3, and it's easy to demonstrate this fact, as reversing that should give the original answer:

(x^3 - a^3) * (x-a) does not equal (x^4 - a^4) when you do an expansion.

Division of polynomials seems like "magic" when you first see the results, I'm guessing you have no idea how you'd go about that. It just seems like you need to memorize the results and just trust they are correct and worked out by previous generations of genius mathematicians. Luckily, it's a lot more simple, but it was never taught to me in class or at college.

Polynomial long division is one way to divide arbitrary polynomials by other arbitrary polynomials. The principles are identical to regular long division, except you use "base x" instead of "base 10". e.g. you need to eliminate the highest power of x on each round, then add those to your final divisor.

First, since x-a has one power of x, and x^4-^4 has 4 powers of x, the (x-a) must be multiplied by x^3 to match the highest term. x^3(x-a) = x^4 - ax^3. Subtract this result from the original polynomial, then add the x^3 factor to your final answer:

x^4 - a^4 - (x^4 - ax^3) = ax^3 - a^4 remainder, with (x^3) times (x-a) taken out.

At this stage it would be perfectly legitimate to write (x^4-a^4)/(x-a) = x^3 + (ax^3 - a^4)/(x-a), which shows how this process works in terms of normal algebra. It's just regrouping the top of the fraction into parts divisible by (x-a) and them eliminating (x-a)/(x-a) for those sections until there's nothing left above a fraction. But to keep it simple, you don't write out the "full equation" each time, you just write out the remainder part that you need to factor next. So we take the remainder (ax^3 - a^4)/(x-a) and do the same process for that as we did for the original fraction

Now, the higest term of x = ax^3, so multiply (x-a) by ax^2 to eliminate the higest term (x-a)*ax^2 = (ax^3 - a^2x^2)

ax^3 - a^4 - (ax^3 - a^2x^2) = a^2x^2 - a^4 remainder, with (x^3 + ax^2) times (x-a) taken out

To eliminate a^2x^2, that's (x-a) times a^2x, i.e. subtract (a^2x^2 - a^3x) :

a^2x^2 - a^4 - (a^2x^2 - a^3x) = a^3x - a^4 remainder, with (x^3 + ax^2  + a^2x) times (x-a) taken out

The final term of x is a^3x, so take out a^3 times (x-a) = (a^3x - a^4)

a^3x - a^4 - (a^3x - a^4), which leaves zero remainder and the job is done.

Now, you take all the factors of (x-a) you needed to remove and add them together, and that's your answer

(x^4-a^4)/(x-a) = (x^3 + ax^2  + a^2x + a^3)

Presto, division of arbitrary polynomials from basic algebraic principles.

[RoguelikeRazuka]

Hey where do I start from if I want to get good at mathematics? I had some calculus, numerical analysis, mathematical physics and functional analysis at university (one term for each of the subjects but calculus), but there was too little time for comprehending either of those subjects (to be honest, I wasn't much hardworking at that point anyway). Now I feel how my mind is increasingly fading away, so it maybe it's already too late for getting smarter (I'm currently 20 y/o)?

[Helgoland]

Dude, I'm twenty as well and only just starting on the really harcore mathematics

I guess you can't 'get good at maths' in that generality. Do you have any concrete applications in mind, or are you interested in the subject abstractly? If the latter, go look into topology - it's a very fun field, especially if you're looking for something a bit more wacky than the 'standard' stuff like algebra or calculus. Start out with point-set topology - all the other kinds build on that. Get a feeling for homotopy equivalences, homeomorphisms, maybe the Euler characteristic and the concept of coverings... Give special attention to the more strange spaces often used as counterexamples, such as the Hawaiian earrings! Those are great fun, and once you've really grasped what makes them special, you know you've gained quite a bit of deeper understanding.

[MagmaMcFry]

If on the other hand you are interested in raw problem solving skills, maybe try stimulating your brain in ways that rewards it when it has a clever idea? What I'm trying to say here is puzzle games. I recommend Portal 2 and Infinifactory for this purpose (I also recommend those games in general). They have great content out of the box, but they also have Steam workshops crammed full of additional, challenging content.

For a more general problem solving experience, you could visit for example this site. It's basically a place to share all kinds of puzzles, logical, mathematical, or otherwise. You can try your hand at those sometime.

[Spehss _]

The derivative of e^(x) is e^(x). Does this apply to any constant^(x) or just the constant e?

Like, would derivative of 2^(x) be 2^(x)?

[crazysheep]

That derivative only applies to the constant e.