The Limit Definition of the Derivative

The limit definition of the derivative computes the slope of a tangent line directly from a function's formula. Write the difference quotient [f(x+h)-f(x)]/h and take the limit as h approaches 0. That limit, when it exists as a finite number, is the derivative at that point. This is the definition from Stewart §2.7, and it is the method you use before you learn any shortcut rules. The derivative from first principles answers one question: what is the instantaneous rate of change of a function at a given input? The answer is that limit.

Every derivative from first principles starts with the difference quotient. For a function f(x), the difference quotient is [f(x+h) - f(x)] / h. The variable h represents a small change in x. The quotient itself is the average rate of change of f over the interval from x to x+h. To get the instantaneous rate of change, take the limit of this quotient as h approaches 0. That limit is the derivative, written f'(x) = lim_{h→0} [f(x+h) - f(x)] / h. The same limit notation appears in Stewart §2.7 as the formal definition. The failure case is when this limit does not exist as a finite number, meaning the function is not differentiable at that point. If the limit is infinite, the derivative does not exist either, even though the function may have a vertical tangent. The technique for calculating limits manually, direct substitution, factoring, rationalizing, applies directly to the difference quotient limit. Check direct substitution first. If it yields 0/0, an indeterminate form, manipulate the expression algebraically before taking the limit.

Apply the Definition

Use f(x) = x², 4x + 1. Write the difference quotient: [f(x+h) - f(x)] / h = [((x+h)², 4(x+h) + 1) - (x², 4x + 1)] / h. Expand: (x² + 2xh + h², 4x, 4h + 1, x² + 4x, 1) / h = (2xh + h², 4h) / h. Factor h: h(2x + h, 4) / h. Cancel h (h ≠ 0 in the limit): 2x + h, 4. Now take the limit as h→0: lim_{h→0} (2x + h, 4) = 2x, 4. The derivative from first principles gives f'(x) = 2x, 4. This matches the power rule you would learn later: derivative of x² is 2x, derivative of -4x is -4, derivative of 1 is 0.

Check at a Point

To find the slope at x = 3, substitute: f'(3) = 2(3)-4 = 2. The tangent line at (3, 2) has slope 2. The limit definition of the derivative works here without needing any other technique. The only algebraic step is expanding the polynomial and cancelling h, which is the standard approach for polynomial functions. If you attempt direct substitution on the original difference quotient, you get 0/0, an indeterminate form. The cancellation resolves it.

Set Up the Difference Quotient

For f(x) = 1/x, the difference quotient is [1/(x+h) - 1/x] / h. Combine the numerator over a common denominator: (x - (x+h)) / (x(x+h)) = -h / (x(x+h)). Then divide by h: [-h / (x(x+h))] / h = -1 / (x(x+h)). The limit as h→0 is lim_{h→0} -1 / (x(x+h)) = -1 / x². So f'(x) = -1/x². This is the derivative from first principles for a rational function. The technique depends on combining fractions, a standard algebraic move. The failure case here is when x = 0: the function is not defined at 0, so the derivative does not exist there either. The limit at 0 is infinite, but as a derivative, it is DNE because no finite slope exists. This example shows why you must check the domain. The difference quotient limit is undefined when x = 0 because the original function is not continuous at 0. Continuity is required for differentiability, though the converse is not true.

Alternative Approach: Common Denominator

If you try direct substitution on [1/(x+h) - 1/x] / h, you get (1/x - 1/x)/0 = 0/0, an indeterminate form. Combining the fractions into a single rational expression is the only manual technique that works here. The same method applies to any rational function: write the numerator as a single fraction, simplify, cancel h, then take the limit.

Rationalize the Numerator

For f(x) = √x, the difference quotient is [√(x+h) - √x] / h. Direct substitution gives 0/0. The technique here is to multiply the numerator and denominator by the conjugate, √(x+h) + √x. This is rationalizing. Multiply: [ (√(x+h) - √x)(√(x+h) + √x) ] / [ h(√(x+h) + √x) ] = [ (x+h) - x ] / [ h(√(x+h) + √x) ] = h / [ h(√(x+h) + √x) ]. Cancel h: 1 / (√(x+h) + √x). Now take the limit as h→0: lim_{h→0} 1 / (√(x+h) + √x) = 1 / (√x + √x) = 1 / (2√x). So f'(x) = 1/(2√x). This is the derivative from first principles for a square root function. The conjugate method is the only way to eliminate the radical in the numerator. The failure case is forgetting the sign when multiplying by the conjugate: (a - b)(a + b) = a² - b², not a² + b².

Domain Check

The derivative 1/(2√x) exists only when x > 0. At x = 0, the difference quotient limit is undefined because the function is not differentiable at an endpoint without a one-sided limit. This matches the fact that √x has a vertical tangent at x = 0, and the derivative is infinite, which is not a finite number. The derivative does not exist at x = 0.

There is an equivalent definition of the derivative at a point a: f'(a) = lim_{x→a} [f(x) - f(a)] / (x - a). This form uses the variable x approaching a instead of h approaching 0. It is useful when the function is given in a way that makes substitution with x easy. For example, with f(x) = 1/x at a = 2, write lim_{x→2} [1/x - 1/2] / (x - 2). Combine the numerator: (2 - x)/(2x) divided by (x - 2) = -(x - 2)/(2x) * 1/(x - 2) = -1/(2x). Then lim_{x→2} -1/(2x) = -1/4. This matches the result from the h-form: f'(2) = -1/4. The alternative form is especially helpful when the function is defined piecewise and you need to check one-sided limits. The difference quotient limit and this form are algebraically identical; choose whichever makes the algebra simpler.