What does h approach in the limit definition of the derivative?

h approaches 0. It never equals 0, because division by 0 is undefined. The limit process evaluates the quotient as h gets arbitrarily close to 0.

Can I use the symmetric difference quotient instead?

The symmetric difference quotient [f(x+h)-f(x-h)]/(2h) approximates the derivative numerically but is not the standard definition. Most textbooks require the standard form with h→0.

Why does h have to cancel before I take the limit?

If h does not cancel algebraically, the quotient becomes a division by 0 at h=0. Cancellation removes the division by 0 and leaves an expression where the limit can be evaluated directly.

What if my function has a constant term like f(x)=x²+5?

The constant cancels when you subtract f(x) from f(x+h). For f(x)=x²+5, f(x+h)-f(x)= (x+h)²+5 - (x²+5) = 2xh+h². The +5 and -5 cancel, leaving only h terms.

Limit Definition of the Derivative (First Principles)

The Limit Definition of the Derivative

The limit definition of the derivative is the single formula that defines every derivative in calculus: f'(x) = lim_{h→0} (f(x+h) - f(x))/h. Asked to use the definition of the derivative? This is the only method that qualifies. It is also called the derivative from first principles. The expression (f(x+h)-f(x))/h is the difference quotient, the slope of the secant line between (x, f(x)) and (x+h, f(x+h)). The derivative is the limit of that slope as h approaches 0, which gives the slope of the tangent line at x.

From Difference Quotient To Derivative

The difference quotient formula [f(x+h)-f(x)]/h converts the average rate of change between two points into the instantaneous rate of change at a single point when you take the limit h→0. This is the definition. OpenStax Calculus Volume 1, section 3.1, and Stewart Calculus, sections 2.7-2.8, both define the derivative as the limit of the difference quotient. The variable h is not a number to solve for; it is an increment that must cancel algebraically from the numerator before the limit is taken. If h does not cancel, the function is not differentiable at that x.

The 5-Step First-Principles Method

Every derivative from first principles follows five steps. Step 1: evaluate f(x+h) by substituting (x+h) for every x in f(x). Step 2: subtract f(x) to get the numerator f(x+h)-f(x). Step 3: simplify the numerator algebraically, combine fractions, expand binomials, or multiply by a conjugate, until every term contains a factor of h. Step 4: divide the entire simplified numerator by h, cancelling that factor. Step 5: take the limit as h→0. The result is f'(x).

Common Error: Premature Division

The most frequent failure is distributing 1/h into f(x+h)-f(x) before combining the numerator. Do this and you break the order of operations. Combine the numerator first, then divide. Paul's Online Math Notes calls this the number-one mistake on his Definition of the Derivative page.

Derivative From First Principles: Polynomial Example

Find the derivative of f(x) = 3x² + 2x + 1 using the limit definition of the derivative. Step 1: f(x+h) = 3(x+h)² + 2(x+h) + 1 = 3(x²+2xh+h²) + 2x+2h + 1 = 3x²+6xh+3h² + 2x+2h + 1. Step 2: f(x+h)-f(x) = (3x²+6xh+3h²+2x+2h+1) - (3x²+2x+1) = 6xh+3h²+2h. Step 3: factor h: h(6x+3h+2). Step 4: divide by h: 6x+3h+2. Step 5: limit as h→0 gives 6x+2. Compare with the power rule: derivative of 3x² is 6x, derivative of 2x is 2, derivative of 1 is 0. Matches.

Find Derivative Using Limit Definition: 1/x Example

For f(x) = 1/x, x≠0. Step 1: f(x+h) = 1/(x+h). Step 2: f(x+h)-f(x) = 1/(x+h) - 1/x. Combine over a common denominator: (x - (x+h)) / (x(x+h)) = -h / (x(x+h)). Step 3: numerator is -h; factor h: h * [-1 / (x(x+h))]. Step 4: divide by h: -1 / (x(x+h)). Step 5: limit as h→0 gives -1 / (x²). Result: f'(x) = -1/x². This matches the power rule: derivative of x⁻¹ is -1x⁻². Do not skip the common denominator step; without it h never cancels.

Definition Of Derivative: Radical Function Example

For f(x) = √x, x≥0. Step 1: f(x+h) = √(x+h). Step 2: f(x+h)-f(x) = √(x+h) - √x. Multiply numerator and denominator by the conjugate: (√(x+h)-√x)(√(x+h)+√x) / (√(x+h)+√x) = (x+h - x) / (√(x+h)+√x) = h / (√(x+h)+√x). Step 4: divide by h: 1 / (√(x+h)+√x). Step 5: limit as h→0 gives 1 / (2√x). This is the derivative of √x. Use the conjugate technique; without it the numerator does not factor h.

The Alternate Form: Limit As X Approaches A

The definition of the derivative also has an alternate form: f'(a) = lim_{x→a} (f(x)-f(a))/(x-a). This is equivalent to the h-form, with x-a playing the role of h. Stewart Calculus (section 2.7) and OpenStax (section 3.1) both present this version. Use it when you know a specific point a and want the derivative there directly, rather than finding a general formula f'(x). The algebra is the same: simplify the numerator until the factor (x-a) cancels.

H Approaches 0: When The Derivative Does Not Exist

The limit definition fails when the one-sided limits of the difference quotient do not agree. The classic example is f(x) = |x| at x=0. For h>0, (|0+h|-|0|)/h = h/h = 1. For h<0, (|0+h|-|0|)/h = (-h)/h = -1. Since the left-hand limit (-1) and right-hand limit (1) differ, lim_{h→0} does not exist. The derivative at 0 does not exist. This failure also occurs at cusps (sharp corners) and at jump discontinuities. Graph the function: at x=0 the absolute value function has a V shape; the secant lines from the left approach slope -1, from the right approach slope +1. No single tangent line exists.

Checking With Derivative Rules

Every derivative computed from first principles can be checked against the standard derivative rules: power rule, product rule, quotient rule, and chain rule. For f(x)=3x²+2x+1, the power rule gives 6x+2, matching the limit result. For 1/x, the power rule gives -1/x². For √x, the power rule gives (1/2)x⁻¹/² = 1/(2√x). If your limit answer does not match the rule, you missed an algebraic step. The rules are shortcuts, not substitutes; they are derived from the limit definition itself. Stewart sections 2.7-2.8 and Paul's Online Math Notes both show this verification.

Practice Set: Derivative From First Principles

Compute each derivative using the limit definition. 1) f(x)=4x-7. 2) f(x)=x²+3x. 3) f(x)=5/(x+2). 4) f(x)=√(x+1). 5) f(x)=|x-2| at x=2. Answers: 1) f'(x)=4. 2) f'(x)=2x+3. 3) f'(x)=-5/(x+2)². 4) f'(x)=1/(2√(x+1)). 5) does not exist at x=2. For each, if your final limit does not match the power-rule or quotient-rule result, check the numerator simplification before the h cancels.