Differentiation from first principals.

blue-army

I'm doing ordinary level....our teacher taught us this using a different formula.....

something like
a = small increase in x
b = small increase in y

anyone else use this method? If so can someone please post an example?

Thanks

MaltEagle

No, that method seems diffucult. Why not use Delta X?


Let y = question asked.

Replace y+delta y (for Y) and x+delta x (for X)

Multiply out the brackets (like Algebra)

Keep going as far as you can

You should get the differenciation of the sum.

Delta is in the shape of a triangle.

Hope this helps.

turgon

of just use the formula - couldn't be simpler!!

y = f(x)

f '(x) = the limit of:
f(x+h)-f(x)
h

as h tends to zero. Usually you just manipulate it a tiny bit to get it out. At ordinary level it shouldn't be that hard. Example

f(x) = 2(x)squared
f(x+h) = 2(x+h)squared
f(x+h) = 2[(x)squared + 2xh + (h)squared]
f(x+h) - f(x) = 2(x)squared + 4xh + 2(h)squared - 2(x)squared
f(x+h) - f(x) = 4xh + 2(h)squared
then divide by h
f(x+h) - f(x) =4x +2h
h
but h=0
f '(x) = 4x

http://skoool.ie/skoool/examcentre_sc.asp?id=716

hatrickpatrick

I found it much easier to do these once I understood what was actually going on:
Essentially, you're trying to find the slope of a tangent to a curve on your graph. Now, you'll probably know the slope formula already from co-ordinate geometry:
y2-y1
x2-x1

Except the problems is, of course, that a tangent only touches the curve once, so you only have one point.
What you have to do is add H to that point (H is a tiny addition to f(x), so small that it doesn't matter). Then you have two points and can find your slope.

So think of it like this:
y1 = f(x)
y2= f(x+h)
x1= x
x2 = (x+h)

Therefore your slope formula is going to be
f(x+h) - f(x)
(x+h)- x

Let's take an example function:
Differentiate x² + 6x - 2
Firstly, check your answer using rule - your derivative is going to be 2x + 6.

Then go through the slope formula, cancelling the equation down until you are left with that answer:

((x+h)² + 6(x+h) - 2) - (x² + 6x - 2)
x + h - x

Don't forget to change the signs in the second bracket since it's -

x² + h² +2xh + 6x + 6h - 2 - x² - 6x + 2
x + h - x

Now you can start cancelling. + 6x will cancel with -6x, + 2 with - 2, and x² with - x², and on the bottom line x cancels with -x leaving you with just h:

h² + 2xh +6h

h

Now you divide h into everything above the line,
h² / h = h,
2xh / h = 2x
6h / h = 6

You're left with:
h + 2x + 6

Now the limit, to get rid of the last h:

lim h -> 0 = 2x + 6

Once you get it into your head that you're just using the slope formula with full expressions instead of numbers the whole thing is ridiculously easy to remember.
N.B - DO NOT FORGET THE LIMIT.
You will lose marks specifically for not mentioning the limit at the end (lim h -> 0)

I know it looks confusing, but trust me it's very easy to remember it once you remember to use the slope formula

SmashingPilot

turgon wrote: »

of just use the formula - couldn't be simpler!!

y = f(x)

f '(x) = the limit of:
f(x+h)-f(x)
h

as h tends to zero. Usually you just manipulate it a tiny bit to get it out. At ordinary level it shouldn't be that hard. Example

f(x) = 2(x)squared
f(x+h) = 2(x+h)squared
f(x+h) = 2[(x)squared + 2xh + (h)squared]
f(x+h) - f(x) = 2(x)squared + 4xh + 2(h)squared - 2(x)squared
f(x+h) - f(x) = 4xh + 2(h)squared
then divide by h
f(x+h) - f(x) =4x +2h
h
but h=0
f '(x) = 4x

http://skoool.ie/skoool/examcentre_sc.asp?id=716

+1 for this. Its the easiest way to remember, I find.

Diarmsquid

Here's y = cosx from first principals.
I typed it out in the Leaving Cert quiz tread, so I said I might as well post it here too. Tricky parts are in bold.

f(x) = cosx, let h = a small change in x
f(x+h) = cos(x+h)

dy/dx = lim(h -> 0) ( f(x + h) - f(x) ) / h

f(x+h) - f(x) = cos(x + h) - cos(x)

cosA - cosB = -2sin((A + / 2).sin((A - / 2)

f(x+h) - f(x) = -2sin(( x + h + x ) / 2).sin(( x + h - x ) / 2)

f(x+h) - f(x) = -2sin(( 2x + h ) / 2).sin( h / 2)

f(x+h) - f(x) / h = -2sin(( 2x + h) / 2).sin( h / 2) / h

(multiply above an below by 1/2)

f(x+h) - f(x) / h = -2sin(( 2x + h) / 2).( sin( h / 2) / (h/2) ). (1/2)

lim (h -> 0)f(x+h) - f(x) / h = -2sin(( 2x + h) / 2).( sin( h / 2) / (h/2) ). (1/2)

(sinx/x = 1)

=-2sin(2x/2).1.(1/2)

= -sinx