Differentiation from 1st principles

**db5vry** · Nov 17th 2008, 01:23 PM

I don't know how to do this but I'd like to learn how:

A question is "Given that 2x² - 5x + 3, find dy/dx from first principles."

Can anyone please help me

**TitaniumX** · Nov 17th 2008, 01:25 PM

Originally Posted by db5vry

I don't know how to do this but I'd like to learn how:

A question is "Given that 2x² - 5x + 3, find dy/dx from first principles."

Can anyone please help me

Hmmmm....

dy/dx = 4x - 5 ?

To learn how to differentiate polynomials.
http://en.wikipedia.org/wiki/Calculus_with_polynomials

**db5vry** · Nov 17th 2008, 01:28 PM

Originally Posted by TitaniumX

Hmmmm....

dy/dx = 4x - 5 ?

To learn how to differentiate polynomials.
Calculus with polynomials - Wikipedia, the free encyclopedia

Yeah I know how to find dy/dx, you just multiply 2 by the ² and get 4 and reduce the power, then minus the 5.
But does anyone know how to do this via first principles?

**TitaniumX** · Nov 17th 2008, 01:35 PM

it's just a fancy way of saying plugging it into the definition of a derivative in terms of the limit as x approaches 0.

**CaptainBlack** · Nov 17th 2008, 01:54 PM

Originally Posted by db5vry

I don't know how to do this but I'd like to learn how:

A question is "Given that 2x² - 5x + 3, find dy/dx from first principles."

Can anyone please help me

Write out the difference quotient for $f(x)=2x^2-5x+3$ :

$<br /> DQ(f,h)=\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)^2-5(x+h)+3-2x^2+5x-3}{h}<br />$

.............. $= \frac{4hx+2h^2-5h}{h}=4x-5+2h$

So:

$<br /> \frac{df}{dx}=\lim_{h \to 0} DQ(f,h)=4x-5<br />$

CB

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