Use the limit theorem definition of a derivative to show that given f(x)=3x^2+1, the derivative f’(x)=6x
(f(x+h)-f(x))/h
= ((3(x+h)^2+1)-(3x^2+1))/h
Now expand all that, divide by h, and take the limit.
Yes, it will come out to be 6x
To use the limit theorem definition of a derivative, we start by considering the function f(x) = 3x^2 + 1 and calculate the difference quotient.
The difference quotient is defined as:
[f(x + h) - f(x)] / h
where h represents a small change in x. We will use this difference quotient to find the derivative of f(x).
Let's begin by replacing f(x) with 3x^2 + 1 in the difference quotient:
[f(x + h) - f(x)] / h
= [3(x + h)^2 + 1 - (3x^2 + 1)] / h
Now we simplify the numerator:
= [3(x^2 + 2xh + h^2) + 1 - 3x^2 - 1] / h
= [3x^2 + 6xh + 3h^2 + 1 - 3x^2 - 1] / h
The terms 3x^2 and -3x^2 cancel out, as well as the constant terms 1 and -1:
= 6xh + 3h^2 / h
Next, we can factor out an h from the numerator:
= h(6x + 3h) / h
Now, we cancel out the h terms in the numerator and denominator:
= 6x + 3h
As h approaches 0, this expression simplifies further. Since the limit of a constant multiplied by h as h approaches 0 is 0, we are left with:
lim[h->0] (6x + 3h)
= 6x + 0
= 6x
Thus, as h approaches 0, the difference quotient simplifies to 6x. This means that the derivative, f'(x), is equal to 6x.
Therefore, we have demonstrated using the limit theorem definition that for the function f(x) = 3x^2 + 1, the derivative f'(x) is equal to 6x.