Differentiation by first principles

Stuffnmusic

Two days from the leaving cert and ive realised i have no idea how to differentiate by first principles, and to make matter worse, I have no notes on it. If anyone could help me out with this id really appreciate it.

sponsoredwalk

[latex]y_0 \ = \ f(x_0)[/latex]
[latex]y_0 \ + \ \Delta \ y \ = \ f(x_0 \ + \ \Delta x )[/latex]
[latex]y_0 \ + \ \Delta y \ - \ y_0 \ = \ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)[/latex]
[latex] \Delta y \ = \ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)[/latex]
[latex] \frac{ \Delta y}{ \Delta x}} \ = \ \frac{ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)}{ \Delta x}[/latex]
[latex] \frac{dy}{dx} \ = \ \lim_{\Delta x \rightarrow 0} \ \frac{ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)}{ \Delta x}[/latex]

So:

[latex]f(x) \ = \ 2x[/latex]
[latex]f(x \ + \ \Delta x) \ = \ 2(x \ + \ \Delta x)[/latex]
[latex]f(x \ + \ \Delta x) \ - \ f(x) = \ 2(x \ + \ \Delta x) \ - \ 2x[/latex]
.
.
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Try [latex]f(x) \ = \ \frac{x}{x \ - \ 1}[/latex].

Stuffnmusic

thanks, i owe you one

TheBody

Check out the following video. It should have everything you need.

http://patrickjmt.com/derivatives-using-the-definition-to-find-a-derivative/

eoins23456

http://www.mathsireland.com/LCHGeneralNotes/Calculus/1stPrin/1st_PrinQuestions.html for higher level

http://www.mathsireland.com/LCPGeneralNotes/calculus/1stPrin/1st_PrinQuestions.html
ordinary level

jody24

differentiate 10-2x with respect to x from first principles

Fish Finger Pie

10

Yakuza

Fish Finger Pie wrote: »

10

Err, newp.

It's almost three years since the OP did their Leaving Cert. Whether they managed to down with calculus or not is fairly moot by now, I'd say.

IBTL.

Fish Finger Pie

So what's the answer then? <snip>

srsly78

The "answer" is to show the method step by step.

Fish Finger Pie

y = 10 - 2x

As x tends to 0, y tends to 10

Therefore, dy/dx = 10.

Right?

srsly78

Right final answer, wrong method. See post #2. It doesn't say x tends to 0, it says delta x tends to 0.

edit: oops not even the right final answer!

Fish Finger Pie

y = 10 - 2x

dy = 10 - 2dx

dy/dx = 10/dx - 2 = 10

sponsoredwalk

Fish Finger Pie wrote: »

y = 10 - 2x

dy = 10 - 2dx

dy/dx = 10/dx - 2 = 10

When you write y = 10 - 2x what you really mean is y(x) = 10 - 2x.
Then y(x + Δx) = 10 - 2(x + Δx).
So y(x + Δx) - y(x) = [10 - 2(x + Δx)] - [10 - 2x] = (10 - 10) - 2x + 2x - 2Δx = - 2Δx.
Then [y(x + Δx) - y(x)]/Δx = - 2.
Taking a limit as Δx ---> 0 gives dy/dx = - 2.

Fish Finger Pie

Where'd the 10 go to?

srsly78

10 - 10 = 0

Michael Collins

This is going nowhere. Locked.