Differentiation help

Spitfir3 · 11 May 2011 at 18:38

Hey guys,

Right, i'm doing some work on differentiation but we've only done two classes on it so far so i'm still on the basics.

I'm having a bit of a trouble with the principles and general ideas I think... for example, could someone just clarify this:

A derivative of a function is the average rate of change of that function ?

I've got a question here, if y = sin x + cos x, find dy/dx.. no problems with that = cos x - sin x... but then it asks me to find the derivative of y at x = pi/2 rads / 90 degrees...

Is that asking for me to find the instantaneous rate of change of the function at the point x = 90 degrees?

Wolfram suggests -1 which is just plugging in 90 for the 2 x's... is it as simple as that?

Also, how does one notate that in an answer? Would it be dy/dx = -1? :confused:

This brings me to my last question, one of the questions is asked:

d/dx ( (cube rt x) + 4 * (fifth rt x) )

Is that just saying y = ( (cube rt x) + 4 * (fifth rt x) ) , find dy/dx of that function?

I can't get my head round that bit :confused:

Cheers guys

Amleto · 11 May 2011 at 18:47

Yes, it is just asking you to 'put 90 in'. Yes, it is simple

Notate it thusly: dy/dx | x=pi/2 = -1
There should be a vertical bar at the right of dy/dx, and at the bottom (like a sub script, write x=pi/2).

The derivative is not 'the average rate of change' it IS the rate of change.

d/dx is differentiate wrt x.

so d/dx( f(x) ) means find f'(x). Or you can think of it as you put it, but stricly, there is no need to bring poor old 'y' in to this

joeyjojo · 11 May 2011 at 19:37

Thumbs up for Amleto's reply.

It's abstract now, but try to think of d/dx as an operator. Which means, it acts on a function of the x variable, and returns you a function of the x variable. Which is why it's correct to write d/dx(loads of stuff) = some other stuff.

Noxia · 11 May 2011 at 19:44

I admire you guys who can do that kind of math, I'm bad at the basics.

GreatAuk · 11 May 2011 at 20:09

Spitfir3 said:
This brings me to my last question, one of the questions is asked:

d/dx ( (cube rt x) + 4 * (fifth rt x) )

Is that just saying y = ( (cube rt x) + 4 * (fifth rt x) ) , find dy/dx of that function?

I can't get my head round that bit

Cheers guys

Yep.

It might help to think of it as (d/dx)(y) rather than dy/dx. You just sub in whatever y is.

Spitfir3 · 11 May 2011 at 21:56

Yes! Excellent, thanks so much guys

Just what I wanted

One thing though, f'(x) - what's the superscript line supposed to notate? First derivative of that function?

Cheers!

Dave · 11 May 2011 at 22:17

Spitfir3 said:
Yes! Excellent, thanks so much guys Just what I wanted

One thing though, f'(x) - what's the superscript line supposed to notate? First derivative of that function?

Cheers!

If I understand the question - f'(x) simply means the first derivative of the function f(x). (so if y = f(x), then f'(x) = dy/dx) Similarly, f''(x) is the second derivative, d^2/dx^2, f'''(x) is the third derivative and so on.

PermaBanned · 11 May 2011 at 22:20

Noxia said:
I admire you guys who can do that kind of math, I'm bad at the basics.

Meh, i presume it's for an A Level maths exam (Core 1/2?). I don't know about the others, but when i do this sort of maths i do it to pass exams, i can't see it having any real world application. That or we just haven't been taught it.

Judgeneo · 11 May 2011 at 22:23

Superewza said:
Meh, i presume it's for an A Level maths exam (Core 1/2?). I don't know about the others, but when i do this sort of maths i do it to pass exams, i can't see it having any real world application. That or we just haven't been taught it.

Plenty of real world applications, try studying some physics, population dynamics, or modelling any sort of data really. Basic calculus comes up everywhere

PermaBanned · 11 May 2011 at 22:29

Judgeneo said:
Plenty of real world applications, try studying some physics, population dynamics, or modelling any sort of data really. Basic calculus comes up everywhere

So why aren't we taught that? I find it really hard to learn things if i can't understand them, i think knowing what they're for is something of an important point for understanding them.

Tikkabhuna · 11 May 2011 at 22:32

I did this 2 years ago and can't remember any of it.

Amp34 · 11 May 2011 at 22:33

When I did my maths A-Level (only a few years ago) there were plenty of "real world" examples, in fact exams usually had real world style questions.

On the other hand I say "real world" (in speech marks) because real world ones are often much harder to understand and work out what to do.

PermaBanned · 11 May 2011 at 22:34

Tikkabhuna said:
I did this 2 years ago and can't remember any of it.

Bring the power to the front and drop it by one? :cool:

Lightnix · 11 May 2011 at 22:34

Superewza said:
So why aren't we taught that? I find it really hard to learn things if i can't understand them, i think knowing what they're for is something of an important point for understanding them.

Check out question 9 of this paper as a simple example of where differentiation can be used to solve a real-world problem:

https://eiewebvip.edexcel.org.uk/Reports/Confidential Documents/0801/6664_01_que_20080109.pdf

Finding the minimum surface area of something like that is clearly applicable in fields such as manufacturing.

Klo · 11 May 2011 at 22:35

You get some differentiation in economics, working out marginal cost curves and stuff.

I hate it though =/

Tikkabhuna · 11 May 2011 at 23:16

Superewza said:
Bring the power to the front and drop it by one?

Ok ok, that's the easy part. I used to remember what sin/cos/tan turned into, etc. I'm going to buy my old core 4 book after I finish reading everything else I have to get through.

p4radox · 11 May 2011 at 23:24

Superewza said:
Meh, i presume it's for an A Level maths exam (Core 1/2?). I don't know about the others, but when i do this sort of maths i do it to pass exams, i can't see it having any real world application. That or we just haven't been taught it.

Calculus is pretty much the foundation of classical and modern physics. It's everywhere.

Didn't you do any mechanics, for instance?

Westyfield2 · 11 May 2011 at 23:27

Sig worthy TBH

Superewza said:
i can't see it having any real world application.

Yes, Calculus has no real world applications :rolleyes:

.

PermaBanned · 11 May 2011 at 23:35

p4radox said:
Calculus is pretty much the foundation of classical and modern physics. It's everywhere.

Didn't you do any mechanics, for instance?

Nope, we've only done D1 and C1 so far, i'll probably do M1 when i get the chance since i'm a lot better at physics than maths. I think i only dropped something like 23 points across the whole Physics GCSE course, don't remember revising for it much either.

clv101 · 11 May 2011 at 23:45

Superewza said:
Meh, i presume it's for an A Level maths exam (Core 1/2?). I don't know about the others, but when i do this sort of maths i do it to pass exams, i can't see it having any real world application. That or we just haven't been taught it.

Pretty much any science/engineering work has A-Level calculus at it's heart... also economics, any quantitative field really. Most things in the real world change at interesting rates to one another, that's calculus.