Exploring the Definition of Derivative: Example Problem and Tangent Line Slope

Definition of Derivative. Example Problem 2

Use the definition of derivative to find f' (x) of the function:
f(x)=3x^2 -4
Let's use the definition of a derivative to find the derivative of the function f where f(x) =
3×2 – 4.
The derivative of a function f(x) is defined as the limit of the difference quotient as h
approaches zero. C| If you recall, the derivative of a function f(x) is defined as lim h-+0 f(x
+ h) – f(x) divided by h.

I will define the function f(x) + h as the modified function f(x), or, for all x's, replacing them
with x plus h.
We can find the limit of f(x) as h approaches zero by evaluating f(x) plus h minus f(x),
where f(x) is given by f(x) = 3×2 – 4.
If you remember from the earlier section, we will get 0/0 if we try to evaluate this limit.
But now the fun really begins.

We can clean all this up by doing some arithmetic. The total
can't be factored, but we can distribute it.
As h approaches 0, the limit is equal to 3. When you multiply x + h times x + h, you get x2
+ 2xh + h2 – 4.
I'm going to distribute the negative sign from the second term to the first term and get -3x?
+ 4. Then I'll divide that by h. It will be a long one.
Thus, as h approaches zero, 3×2+6xh+3h2-4-3×2+4 divided by h.
Here we have 3×2-3×2, which equals 0.

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We then have 4-4, which also equals 0. And we
have an h left over that I will factor out.
So, when h approaches 0, the limit becomes 6x plus 3h divided by 1.
After all that work, we're left with just 6x divided by 3h.
Letting h = 3x, we find that 3h + 3 • 0 = 6h. Hence, the derivative of x is 6x.
If the question had asked, for example, what is the slope of the tangent line when x equals
2. then we would proceed as follows.
When x equals 2, the derivative of f(x) is 12 if f(2) is used in the formula. If the function f(x)
equals two then its derivative, f(x), is twelve. Negative two – negative twelve. And so on.
f(x)=3x^3 -4
f(x+h)=3(x+h)^2 -4
lim_(h->0) (f(x+h)-f(x))/h
lim_(h->0) (3(x+h)^2 -4-(3x^2 -4))/h
lim_(h->0) ((3(x^2 +2xh+h^2))-4-3x^2 +4)/h
lim_(h->0) (3x^2 +6xh+3h^2 -4-3x^2 +4)/h
lim_(h->0) (h(6x+3h))/h
lim_(h->0) 6x+3h=6x=f'(x)
x=2, f'(2)=12