How does
Limit as delta x approaches 0 =
4x + 2deltax +1 simplify to 4x +1?
I'm supposed to find the derivative using the limiting process
orig equation:
f(x) 2x^2+x-1
The term
2 delta x
vanishes when you take the limit of delta x to zero.
To find the derivative of a function using the limiting process, you need to compute the limit of the difference quotient as "delta x" approaches 0.
Let's start with the original equation:
f(x) = 2x^2 + x - 1
To find the derivative, we need to evaluate the limit as "delta x" approaches 0 of the difference quotient: (f(x + delta x) - f(x))/delta x.
Let's substitute the original equation into the difference quotient:
(f(x + delta x) - f(x))/delta x = [2(x + delta x)^2 + (x + delta x) - 1 - (2x^2 + x - 1)]/delta x
Now, expand and simplify the equation step by step:
1. Distribute the squared term:
= [2(x^2 + 2x * delta x + (delta x)^2) + (x + delta x) - 1 - (2x^2 + x - 1)]/delta x
2. Combine like terms:
= [2x^2 + 4x * delta x + 2(delta x)^2 + x + delta x - 1 - 2x^2 - x + 1]/delta x
3. Cancel out common terms:
= [4x * delta x + 2(delta x)^2 + delta x]/delta x
4. Divide both the numerator and denominator by delta x:
= 4x + 2(delta x) + 1
Now, let's take the limit as "delta x" approaches 0:
lim(delta x -> 0) [4x + 2(delta x) + 1]
As "delta x" approaches 0, the term 2(delta x) approaches 0. Therefore, our final result is:
lim(delta x -> 0) [4x + 1] = 4x + 1
Hence, the simplified derivative of the function f(x) = 2x^2 + x - 1 is 4x + 1.