find from first principle,the derivative of 2x^2-x with respect to x
y(x) = 2 x^2 - x
y (x+h) = 2 (x+h)^2 - x - h
so
y(x+h) = 2 x^2 + 4 x h + 2 h^2 - x - h
and
y(x+h)-y(x) = 4 x h + 2 h^2
divide by h
[y(x+h)-y(x)] / h = 4 x + 2 h
let h ---> 0
4 x
TYPO !
y(x) = 2 x^2 - x
y (x+h) = 2 (x+h)^2 - x - h
so
y(x+h) = 2 x^2 + 4 x h + 2 h^2 - x - h
and
y(x+h)-y(x) = 4 x h + 2 h^2 - h
divide by h
[y(x+h)-y(x)] / h = 4 x + 2 h -1
let h ---> 0
4 x - 1
To find the derivative of the function f(x) = 2x^2 - x using first principles, we need to apply the definition of the derivative.
The derivative of a function is defined as the limit of the difference quotient as the change in x approaches zero.
Let's start by applying the difference quotient formula:
f'(x) = lim(h->0) [(f(x+h) - f(x))/h]
Replacing f(x) with 2x^2 - x, we have:
f'(x) = lim(h->0) [(2(x+h)^2 - (x+h) - (2x^2 - x))/h]
Expanding the expression inside the limit:
f'(x) = lim(h->0) [(2(x^2 + 2xh + h^2) - x - h - 2x^2 + x)/h]
Simplifying the expression:
f'(x) = lim(h->0) [2x^2 + 4xh + 2h^2 - x - h - 2x^2 + x)/h]
Canceling out the 2x^2 and x terms:
f'(x) = lim(h->0) [4xh + 2h^2 - h)/h]
Dividing each term by h:
f'(x) = lim(h->0) [4x + 2h - 1]
Taking the limit as h approaches zero, the h term vanishes:
f'(x) = 4x - 1
Therefore, the derivative of f(x) = 2x^2 - x, using the first principles, is f'(x) = 4x - 1.
To find the derivative of a function using first principles, we need to use the definition of the derivative.
The definition of the derivative of a function f(x) at a point x=a is given by:
f'(a) = lim(h->0) [f(a+h) - f(a)] / h
Now, let's find the derivative of the function f(x) = 2x^2 - x using this definition.
1. Replace f(x) with the given function: f(x) = 2x^2 - x.
2. Replace f(a) with the function evaluated at a: f(a) = 2a^2 - a.
3. Replace f(a+h) with the function evaluated at a+h: f(a+h) = 2(a+h)^2 - (a+h).
4. Substitute these values into the definition of the derivative:
f'(a) = lim(h->0) [f(a+h) - f(a)] / h
= lim(h->0) [(2(a+h)^2 - (a+h)) - (2a^2 - a)] / h
= lim(h->0) [2(a^2 + 2ah + h^2) - a - h - 2a^2 + a] / h
= lim(h->0) [2a^2 + 4ah + 2h^2 - a - h - 2a^2 + a] / h
= lim(h->0) [4ah + 2h^2 - h] / h
= lim(h->0) 4a + 2h - 1
5. Take the limit as h approaches 0:
f'(a) = 4a + 2(0) - 1
= 4a - 1
Therefore, the derivative of the function f(x) = 2x^2 - x with respect to x is f'(x) = 4x - 1.