# Thread: Differentiation from 1st principles

1. ## Differentiation from 1st principles

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."

2. 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."

Hmmmm....

dy/dx = 4x - 5 ?

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

3. 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?

4. 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.

5. 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."

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

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

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

So:

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

CB