
Topic review (newest first)
 Ricky
 20051215 11:55:51
 Tredici
 20051215 11:07:11
3? Out of 3? Correct? .
Lmao, that's crazy, wouldn't of done it without you guys.
Let's suppose the last question was:
f(x) = 3e^x  ½ ln (x  2) as oppose to f(x) = 3e^x  ½ ln x  2
Would it then be..
(3e^x)' = 3e^x
(½ ln x  2)' = is that just, ½ 1/(x2)? or simplified 1/(2x4)
∴ ƒ'(x) = 3e^x  1/(2x4). Not quite so sure on that one, to be honest.
 Ricky
 20051215 10:13:10
3/3, well done. They are all correct.
 Tredici
 20051215 09:48:43
Ok, and I'm back again.
ƒ(x) = 3 sin ² x + sec 2x
Ok, I'm not sure if this is correct, but I made that out to be:
(3 sin ² x)' = using chain rule, 3 [ 2 (sin x) (cos x) ] = 6 sin x cos x
(sec 2x)' = 2 sec (2x) tan (2x)
∴ ƒ'(x) = 6 sin x cos x + 2 sec (2x) tan (2x)
Please correct me if I'm wrong .
Another question.
[x + ln(2x)] ³
I'm very unsure of whether this is correct, but I ended up with:
[x + ln(2x)]' = (1 + 1/x)
using chainrule I ended up with:
ƒ'(x) = 3 [x + ln(2x)] ² (1 + 1/x)
Ok, looking back on that, it looks very wrong. So corrections would never go a miss.
Final One:
f(x) = 3e^x  ½ ln x  2
Hmm... let's begin
(3e^x)' = 3e^x, right?
(½ ln x)' = ½ 1/x = 1/2x, right?
(2)' = 0
∴ ƒ'(x) = 3e^x  1/2x.
Hmmm.. please let me know how wrong I am on these, lol. Only way to improve, eh?
Thanks again guys, as I've said before, really appreciate what you guys do, I'm surprised you don't charge!
 mathsyperson
 20051215 09:06:59
You don't have to use 't', that's just how we were taught. t is just a function of x, chosen to make the expression easier to differentiate.
 Tredici
 20051215 09:06:51
Oh guys I've just noticed, I haven't made myself too clear; I've written a x sin(x) when I meant more along the lines of (a)times[sin(x)]. How would that be differenciated?
Edit: Don't worry, I got it . Silly me.
 Tredici
 20051215 09:01:43
 mathsyperson
 20051215 05:59:38
Indeed I be. Yarr.
ryos wrote:(sin²(x))′ = sinxcosx + cosxsinx = 2sinxcosx (there's probably and identity to simplify that further, but I can never remember those darn things).
2 sinxcosx ≡ sin 2x
Arrrgh, mathsy be a quick one.
sin(ax)
The chain rule is most convenient for this one. (If you haven't learned it, the chain rule is a rule for differentiating composed functions. If ƒ(x) = sinx and g(x) = ax, then sin(ax) = ƒ(g(x)). The chain rule goes like this: (ƒ(g(x)))′ = ƒ(g(x))′ * g′(x).)
So. (sinx)′ = cosx, and (ax)′ = a. Plug these into the chain rule, and out pops a*cos(ax).
The same principle applies to the rest of the functions in that list. Another example is (ln(ax))′ = a/ax = 1/x.
sin²(x)
Rewrite this one as (sinx)(sinx). You can then use the product rule ((ƒ(x)g(x))′ = ƒ(x)g′(x) + ƒ′(x)g(x)).
(sin²(x))′ = sinxcosx + cosxsinx = 2sinxcosx (there's probably and identity to simplify that further, but I can never remember those darn things).
Again, the same idea applies to the rest of the list.
ax*sin(x) Product rule again. ax*cosx + a*sinx. Or did you mean to say a*sin(x)? That's just a*cos(x). Remember that (aƒ(x))′ = a(ƒ′(x))
a x sin(bx + c)
This one's an application of chain and product rules both. sin(bx + c)′ = b*cos(bx + c) (ax*sin(bx+c))′ = abx*cos(bx+c) + a*sin(bx+c)
I hope that helps you. If you were looking more for explanations of why the various rules works, there are some good proofs in my calculus book...
 mathsyperson
 20051215 04:59:52
Each of the sections use the same method, so I'll show you how to differentiate the first of each one.
f(x) = sin (ax)
For this one, you need to use the chain rule: dy/dx = dy/dt * dt/dx. Substitute t = ax.
dy/dt = cos t, dt/dx = a, ∴ dy/dx = a cos t = a cos (ax) You'll find that that result is the same for most of them, but not all, so be careful. For example, ln (ax) differentiated is still 1/x.  I'll come back to sin² x, because it's trickier than the rest.
a sin x, by contrast, is easy. Just take out the constant a and differentiate as normal, before putting the constant back on.
d(a sin x)/dx = a * d(sin x)/dx = a cos x.  a sin (bx + c) is just a combination of the above two things, and you should get a result of ab cos (bx + c).  As I said, sin² x is a bit trickier. For things like this, you need the product rule. d(uv)/dx, where u and v are functions of x, = u*dv/dx + v*du/dx.
In this case, u = v = sin x.
d(sin² x)/dx = 2*sinx*d(sinx)/dx = 2sin x cos x.  I hope I helped.
 Tredici
 20051215 03:31:21
I've had a good look at all the existing topics based around this rather annoying subject, but nothing seems to satisfyingly relate to what I'm questioning. I don't quite understand how special functions react to differentiation, well I suppose I do, if basic differentiation. I'm just really having problems with questions involving the differentiation of multiple functions that are multiplied or squared.
There's a heck of a lot of questions I'm having problems with, and I'd much rather learn how to solve them as oppose to watching someone else work through it for me. So, I was thinking, would it be possible if someone could fill out a list of how the functions are differentiated, maybe if I were to provide a structure? At least, I know this:
f(x) f'(x)
sin(x) cos(x) cos(x) sin(x) tan(x) sec²(x) sec(x) sec(x)tan(x) cosec(x) cosec(x)cot(x) cot(x) cosec²(x) ln(x) 1/x e^x e^x
I know, it's not much at all, but I'm just quite unsure of the following:
f(x) f'(x)
sin(ax) cos(ax) tan(ax) sec(ax) cosec(ax) cot(ax) ln(ax) e^(ax)
f(x) f'(x)
sin²(x) cos²(x) tan²(x) sec²(x) cosec²(x) cot²(x) ln²(x)
f(x) f'(x)
a x sin(x) a x cos(x) a x tan(x) a x sec(x) a x cosec(x) a x cot(x) a x ln(x) a x e^(x)
and finally, just as an example:
f(x) f'(x)
a x sin(bx + c) a x cos(bx + c) a x tan(bx + c) a x sec(bx + c) a x cosec(bx + c) a x cot(bx + c) a x ln(bx + c) a x e^(bx + c)
I'd be eternally greatful if someone could show me how the above are differentiated. I understand it's an extensive list, maybe excessive, so it's perfectly understandable if you don't have the time for me, but I'm just eager to get to grips with this, as to cease resorting to this comunity with similar questions. Thanks so much for your time, I really do appreciate it. David.
