zaqwerty
Member
Do we need to know how to differentiate by first principles? Got told way back in year 11 that we'd need to know it for the HSC but I've never seen any of those types of questions in the exams that I've done...?
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
First principles? (1 Viewer)
- Thread starter zaqwerty
- Start date
I learnt it way back at the start of yr 11 extension havnt seen it since. I think for the first time in 20 years it appeared in the extension paper in 05??
The chance of it being in the 2 unit paper is minimal.
The chance of it being in the 2 unit paper is minimal.
watatank
=)
it's part of the syllabus so you *can* be examined on it, but chances are you won't. all the same learn it.
Forbidden.
Banned
There was a question on a certain dot point in Chemistry that didn't appear from the 2001 paper and it appeared this year.
I would still remember how to differentiate via first principles anyway.
I would still remember how to differentiate via first principles anyway.
Last edited:
- Joined
- Feb 16, 2005
- Messages
- 8,390
- Gender
- Male
- HSC
- 2006
It's really easy to apply it if you understand it. It's easy to remember the formula if you know where it came from.
Basically the aim is to find the gradient of a curve.
A good approximation of the gradient is to plot a secant/straight line through the curve and find the gradient of the straight line. As we approach closer and closer to the actual gradient at a point, the approximate gradient approaches closer to the actual gradient.
So say we have y = f(x), the usual straight line formula for a gradient is (y<sub>2</sub> - y<sub>1</sub>) / (x<sub>2</sub> - x<sub>1</sub>)
If we replace y with f(x), we get:
[f(x<sub>2</sub>) - f(x<sub>1</sub>)] / [x<sub>2</sub> - x<sub>1</sub>]
If x<sub>2</sub> and x<sub>1</sub> are really close to each other then we can let Δx (or more often called h) = x<sub>2</sub> - x<sub>1</sub>
Which implies x<sub>2</sub> = x<sub>1</sub> + Δx
so the approximation is [f(x<sub>1</sub> + Δx) - f(x<sub>1</sub>)] / Δx
As x<sub>2</sub> and x<sub>1</sub> get really close to each other (almost to where they're equal but never exactly equal), then Δx which approach zero, hence the first principles formula (with generalisation of x<sub>1</sub> to just the variable x) is;
lim<sub>Δx --> 0</sub> [f(x + Δx) - f(x)] / Δx
Basically it's just:
lim<sub>'a little bit' --> 0</sub> [f(x + 'a little bit') - f(x)] / 'a little bit'
Basically the aim is to find the gradient of a curve.
A good approximation of the gradient is to plot a secant/straight line through the curve and find the gradient of the straight line. As we approach closer and closer to the actual gradient at a point, the approximate gradient approaches closer to the actual gradient.
So say we have y = f(x), the usual straight line formula for a gradient is (y<sub>2</sub> - y<sub>1</sub>) / (x<sub>2</sub> - x<sub>1</sub>)
If we replace y with f(x), we get:
[f(x<sub>2</sub>) - f(x<sub>1</sub>)] / [x<sub>2</sub> - x<sub>1</sub>]
If x<sub>2</sub> and x<sub>1</sub> are really close to each other then we can let Δx (or more often called h) = x<sub>2</sub> - x<sub>1</sub>
Which implies x<sub>2</sub> = x<sub>1</sub> + Δx
so the approximation is [f(x<sub>1</sub> + Δx) - f(x<sub>1</sub>)] / Δx
As x<sub>2</sub> and x<sub>1</sub> get really close to each other (almost to where they're equal but never exactly equal), then Δx which approach zero, hence the first principles formula (with generalisation of x<sub>1</sub> to just the variable x) is;
lim<sub>Δx --> 0</sub> [f(x + Δx) - f(x)] / Δx
Basically it's just:
lim<sub>'a little bit' --> 0</sub> [f(x + 'a little bit') - f(x)] / 'a little bit'
Last edited:
Starcraftmazter
Member
You should know it, since it uses limits, which are used by about 100 other things in mathematics.