From Average to Instantaneous: The Limit Definition of the Derivative

Secant Slopes and the Need for a Limit

To measure how fast a function changes at a single input, we start with something we can compute: the slope between two points on the graph. Pick a point x and a nearby point x+h (where h is a nonzero change in input). The slope of the secant line through (x,f(x)) and (x+h,f(x+h)) is

m_secant = (f(x+h)-f(x))/((x+h)-x) = (f(x+h)-f(x))/h

This expression is called the difference quotient. It gives the average rate of change over the interval from x to x+h.

From “average over an interval” to “instantaneous at a point”

If we choose smaller and smaller values of h, the second point x+h moves closer to x. The secant line rotates toward a limiting position: the tangent line at x. The derivative is defined as that limiting slope:

f'(x) = lim_{h→0} (f(x+h)-f(x))/h

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Important details:

• h cannot equal 0 inside the fraction. The difference quotient divides by h, so it is undefined at h=0. We simplify algebraically for h≠0, and only then take the limit as h→0.
• Domain restrictions matter. The expressions f(x) and f(x+h) must both be defined. That can restrict which x values are allowed and which small h values are allowed while approaching 0.

Example 1: f(x)=x^2 (Algebraic Simplification)

Start with the difference quotient:

f'(x) = lim_{h→0} [(x+h)^2 - x^2]/h

Expand and simplify (keeping h≠0 during cancellation):

[(x+h)^2 - x^2]/h = [x^2 + 2xh + h^2 - x^2]/h = [2xh + h^2]/h

Factor out h in the numerator:

[2xh + h^2]/h = [h(2x + h)]/h = 2x + h   (valid only for h≠0)

Now take the limit:

f'(x) = lim_{h→0} (2x + h) = 2x

Derivative at a point (example)

At x=3, f'(3)=2·3=6. This means the tangent slope at x=3 is 6.

Example 2: f(x)=1/x (Fractions and Domain Restrictions)

First note the domain: f(x)=1/x is defined only for x≠0. Also, f(x+h)=1/(x+h) requires x+h≠0. While taking h→0, we consider small h values that keep x+h nonzero.

f'(x) = lim_{h→0} [1/(x+h) - 1/x]/h

Combine the fractions in the numerator:

1/(x+h) - 1/x = [x - (x+h)]/[x(x+h)] = [-h]/[x(x+h)]

Substitute back into the quotient:

[1/(x+h) - 1/x]/h = {[-h]/[x(x+h)]}/h = [-h]/[x(x+h)h]

Cancel h (again, only because h≠0 during simplification):

[-h]/[x(x+h)h] = -1/[x(x+h)]

Now take the limit:

f'(x) = lim_{h→0} -1/[x(x+h)] = -1/[x·x] = -1/x^2

Derivative at a point (example)

At x=2, f'(2)=-1/4. The negative sign indicates the graph is decreasing there.

Example 3: f(x)=√x (Radicals and Rationalizing)

Domain: √x is defined for x≥0. Also √(x+h) requires x+h≥0. When taking h→0, we restrict to small h that keep x+h nonnegative.

f'(x) = lim_{h→0} [√(x+h) - √x]/h

This form is not directly simplifiable by cancellation. Multiply by the conjugate to rationalize:

[√(x+h) - √x]/h · [√(x+h) + √x]/[√(x+h) + √x]

Use the difference of squares:

(√(x+h) - √x)(√(x+h) + √x) = (x+h) - x = h

So the expression becomes:

[h] / [h(√(x+h) + √x)]

Cancel h (valid for h≠0):

1 / (√(x+h) + √x)

Now take the limit:

f'(x) = lim_{h→0} 1/(√(x+h) + √x) = 1/(√x + √x) = 1/(2√x)

Important domain note for the derivative

f'(x)=1/(2√x) requires x>0 (because the denominator cannot be 0). So √x is defined at x=0, but its derivative formula does not apply at x=0.

Derivative at a point (example)

At x=4, f'(4)=1/(2·2)=1/4.

Why We Keep Saying “h Cannot Be 0”

The derivative definition uses a limit, not direct substitution. The expression

(f(x+h)-f(x))/h

is undefined at h=0. The strategy is:

• Work with the expression for h≠0 and simplify it.
• After simplification removes the problematic division (often by cancellation), take the limit as h→0.

In the examples above, cancellation of h was legitimate only because we never set h=0 during algebraic steps.

Interpreting the Sign and Steepness of f'(x)

Once you have f'(x), you can check whether it matches the graph’s behavior:

• If f'(x)>0, the function is increasing at x (tangent slopes upward to the right).
• If f'(x)<0, the function is decreasing at x.
• If |f'(x)| is larger, the graph is steeper at that point; if smaller, it is flatter.

Practice Set (Structured)

A. Compute the derivative at a point using the limit definition

• 1) f(x)=x^2: compute f'(5) using lim_{h→0}[(5+h)^2-25]/h.
• 2) f(x)=1/x: compute f'(1) using lim_{h→0}[1/(1+h)-1]/h. State any restrictions needed while simplifying.
• 3) f(x)=√x: compute f'(9) using lim_{h→0}[√(9+h)-3]/h, rationalizing as needed.

B. Find the derivative function f'(x) using the limit definition

• 4) f(x)=x^2: derive f'(x) from lim_{h→0}[(x+h)^2-x^2]/h.
• 5) f(x)=1/x: derive f'(x) from lim_{h→0}[1/(x+h)-1/x]/h, and state the domain of f and f'.
• 6) f(x)=√x: derive f'(x) from lim_{h→0}[√(x+h)-√x]/h, and state where the derivative formula is valid.

C. Verify by comparing with graph behavior (sign and relative steepness)

• 7) For f(x)=x^2, compare f'(-2), f'(0), and f'(3). Which point has the steepest tangent? Which direction (increasing/decreasing) at each point?
• 8) For f(x)=1/x, compare the steepness at x=-1 versus x=-4 using |f'(x)|. Which is steeper, and does that match the graph near those points?
• 9) For f(x)=√x, compare f'(1), f'(4), and f'(16). Explain how the values support the idea that the graph flattens as x increases.
• 10) Domain check: For each function (x^2, 1/x, √x), list the allowed x values for f(x) and for f'(x). Identify any x where f exists but f' does not.