keeperofdakeys wrote:Now try this:
1. d/dx(x*y^{2} + tan(32/x)) =
(hint: d/dx(tan(x)) = sec^{2}(x) = 1/cos^{2}(x) )
2. d/dx( y*ln(2*y^{3}*x) ) =
Are you assuming y is a function of x or that y is a constant?
Moderators: gmalivuk, Moderators General, Prelates
keeperofdakeys wrote:Now try this:
1. d/dx(x*y^{2} + tan(32/x)) =
(hint: d/dx(tan(x)) = sec^{2}(x) = 1/cos^{2}(x) )
2. d/dx( y*ln(2*y^{3}*x) ) =
keeperofdakeys wrote:1. d/dx(x*y^{2} + tan(32/x)) =
(hint: d/dx(tan(x)) = sec^{2}(x) = 1/cos^{2}(x) )
keeperofdakeys wrote:2. d/dx( y*ln(2*y^{3}*x) ) =
vilidice wrote:I'm going to try to take a stab at the sin(x) -> cos(x) thing that you're having problems with (I was always dissatisfied with the way calculus was taught so I've been trying to figure out an elementary way of explaining stuff like this in an elementary way).
The first big thing I need to explain is summations and functions.
The general idea is that some functions and limits can be described using an infinite order polynomyal (there are other kinds too but we'll stick to simple stuff now). The most common introduction to this is the exponential function:
[math]e^x[/math] which is approximated by the series (infinite sum): [math]\sum_{n=0}^{\infty} \frac{1}{n!} X^n = 1+\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6}+...[/math] which is clearly a polynomial (which can be differentiated as such)
tckthomas wrote:
First, thanks!
I should give myself a test, then tell me what I have left to learn
d/dx ln(x) = 1/x
and you can convert any base log into natural log: base a log x = base e log x / base e log a
there seems to be too much trig identities here! i can only remember the reciprocal ones, sine(x) = 1/cosecant(x), cosine(x) = 1/secant(x), tangent(x) = 1/cotangent(x)
this i need more experience, this is the reason why Calculus in 20 minutes fails.
factorials? i think they are just too weird. is it impossible to derive a factorial if it isn't over another factorial?
implicit differentiation i think i get it. derive both sides with respect to x, then dy/dx magically appears, then put that on one side and done.
chain rule/product rule/quotient rule, i need to find out when to use the chain rule. I just don't know why to use the chain rule when sin(3x^2+1)... can you tell me more?
squeeze theorem, heh this is quite cool! squeezing functions into a point... its true when there is upper bound and lower bound functions and when they are equal to something, the middle is squeezed to also equal that.
spins on calculus, making it to a series, AAAHHHHH! freaky stuff!
thanks! this will help me the next 2 years when i do additional math. i would learn calculus secondary 3 or 4...
EDIT: yeah, either 0, -infinity, or infinity, 3 values...
tckthomas wrote:AHHH! I really don't know!
keeperofdakeys wrote:1. d/dx(x*y^{2} + tan(32/x)) =
(hint: d/dx(tan(x)) = sec^{2}(x) = 1/cos^{2}(x) )
tckthomas wrote:Here I go again!keeperofdakeys wrote:1. d/dx(x*y^{2} + tan(32/x)) =
(hint: d/dx(tan(x)) = sec^{2}(x) = 1/cos^{2}(x) )
d/dx(x*y^{2})+d/dx(tan(32/x))
y^{2}+2xy(dy/dx)+sec^{2}(32/x) * d/dx(32/x)
y^{2}+2xy(dy/dx)+sec^{2}(32/x) * -32/x^2
y^{2}+2xy(dy/dx)-(32sec^{2}(32/x))/x^2
I think i got it right!
Edit: no, wolfram alpha didn't lie to me, I didn't use it. I just didn't use the product rule.
314man wrote:You can't derive factorials, and it doesn't come up too often with integral and derivative calculus (at least that I found so far... I'm in 2nd year actuarial science so I'm on multivariable calculus right now). But it comes up quite often when you get into series. Pretty much what happens is factorials pop up, it's a big pain in the ass, so you want to get rid of it. So essentially you want to know how to cancel out factorials. There are the easier ways like n! / (n-1)! = n. But there are some that kind of go against logic. For example, when n goes to infinite, n! / n^n is 1. You can break it down by ((n)(n-1)(n-2)(n-3)...etc.) / (n * n * n * n * etc.). Essentially all the terms have a limit of 1 so the whole thing goes to one. The reason I think it goes againt logic because personally, I think it should go to 0. Because the first term is 1/n, (the 1 being the first number in the factorial). And that limit is 0. And the rest of the terms is less than 1, so it isn't growing so I think it should go to 0. Does anyone want to help me out explaining this to me? (wooh asking help when I'm trying to help )
There are quite a bit of trig identities, but honestly I forgot most of them, but there are a few ones to know that are easy to see and it'll make your life a whole lot easier.
cos^2(x) + sin^2(x) = 1
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2cos^2(x) = 2sin^2(x) + 1 ** (I haven't used this one in a while and I might've made a mistake. It might be the signs are reverse so double check on me before using this one. Also it is an identity to both because to switch from one to the other, you just use the cos^2(x) + sin^2(x) = 1 indentity)
sin(x + y) = ?? (I forgot... I think it is (sin(x/2) + sin(y/2)) / 2 but my memory is a little iffy
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l wrote:314man wrote:You can't derive factorials, and it doesn't come up too often with integral and derivative calculus (at least that I found so far... I'm in 2nd year actuarial science so I'm on multivariable calculus right now). But it comes up quite often when you get into series. Pretty much what happens is factorials pop up, it's a big pain in the ass, so you want to get rid of it. So essentially you want to know how to cancel out factorials. There are the easier ways like n! / (n-1)! = n. But there are some that kind of go against logic. For example, when n goes to infinite, n! / n^n is 1. You can break it down by ((n)(n-1)(n-2)(n-3)...etc.) / (n * n * n * n * etc.). Essentially all the terms have a limit of 1 so the whole thing goes to one. The reason I think it goes againt logic because personally, I think it should go to 0. Because the first term is 1/n, (the 1 being the first number in the factorial). And that limit is 0. And the rest of the terms is less than 1, so it isn't growing so I think it should go to 0. Does anyone want to help me out explaining this to me? (wooh asking help when I'm trying to help )
Generally the factorials that appear in calculus are constants, so of course you can derive them, just like any other constant. Their derivative is 0. In some more advanced stuff you'll get to where the factorial is not a constant, but a function of x, but then we need to replace the factorial (which is only defined on the natural numbers) with something defined on the real numbers, and the definition makes the derivative not so bad.
Now, as for n!/n^n... that's definitely 0 and is a simple application of Stirling's Approximation, which shows that as n gets large, the terms inside the limit are approximately
[math]\frac{\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n}{n^n} =\frac {\sqrt{2 \pi n}}{e^n}[/math]
and [imath]e^n[/imath] grows waaaay faster than [imath]\sqrt n[/imath] so it's 0.
Your argument only shows that nPr/n^r goes to 1. It doesn't work for n! because while each quotient goes to 1, you keep adding more and more smaller quotients. In fact, the argument you gave for why it should be 0 is exactly right. You showed that the limit grows is smaller than 1/n, and it's obviously greater than 0, so by the squeeze theorem it must be 0.There are quite a bit of trig identities, but honestly I forgot most of them, but there are a few ones to know that are easy to see and it'll make your life a whole lot easier.
cos^2(x) + sin^2(x) = 1
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2cos^2(x) = 2sin^2(x) + 1 ** (I haven't used this one in a while and I might've made a mistake. It might be the signs are reverse so double check on me before using this one. Also it is an identity to both because to switch from one to the other, you just use the cos^2(x) + sin^2(x) = 1 indentity)
sin(x + y) = ?? (I forgot... I think it is (sin(x/2) + sin(y/2)) / 2 but my memory is a little iffy
You can get all the addition formulae quickly from expanding out [imath]e^{i(a+b)}=e^{ia}e^{ib}[/imath] using [imath]e^{ix}=\cos x + i \sin x[/imath].
Anyway, cos(2x) = 2cos^2 x - 1=1-2sin^2 x. The missing ones are
sin (x+y) = sin x cos y + sin y cos x
cos (x+y) = cos x cos y - sin x sin y
For cases like this you can do some simple error checking based on other facts you know. For example, cos(2x) couldn't possibly be 2sin^2 x + 1, since the first summand is nonnegative, this value is always at least than 1, and cos(2x) is at most 1. If you have a guess for what sin (x+y) is, you can plug in x=y to see if it agrees with what you know sin (2x) to be.
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