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The Derivative

The derivative is the limit of the secant gradient as the gap between two points shrinks to zero.

Defining the Derivative

f'(x) = lim(h\to 0) [f(x + h) - f(x)] \div h This is exactly the secant-gradient idea from two lessons ago, written as a limit. Instead of shrinking h with a table, algebra lets you find the exact limit in one go.

Example 1 — deriving f'(x) for f(x) = x²

Use the definition to find the derivative of f(x) = x^{2} . 1 f(x + h) = (x + h)² = x^{2} + 2 xh + h^{2} 2 f(x + h) - f(x) = (x^{2} + 2 xh + h^{2}) - x^{2} = 2 xh + h^{2} 3 Divide by h: (2xh + h^{2}) \div h = 2 x + h 4 Let h approach 0 : 2 x + h \to 2 x Answer: f'(x) = 2 x

Example 2 — checking against earlier numbers

Use f'(x) = 2 x to find the gradient of y = x^{2} at x = 3 . 1 f'(3) = 2 \times 3 = 6 This matches the shrinking-h table from two lessons ago, where the secant gradient at x = 3 was heading straight for 6 .

Practice

Using f'(x) = 2 x for f(x) = x^{2}, find f'(4).

Your answer

More lessons in Calculus · Next: The Power Rule · Previous: Limits

Defining the Derivative

f'(x) = lim(h\to 0) [f(x + h) - f(x)] \div h This is exactly the secant-gradient idea from two lessons ago, written as a limit. Instead of shrinking h with a table, algebra lets you find the exact limit in one go.

Example 1 — deriving f'(x) for f(x) = x²

Use the definition to find the derivative of f(x) = x^{2} . 1 f(x + h) = (x + h)² = x^{2} + 2 xh + h^{2} 2 f(x + h) - f(x) = (x^{2} + 2 xh + h^{2}) - x^{2} = 2 xh + h^{2} 3 Divide by h: (2xh + h^{2}) \div h = 2 x + h 4 Let h approach 0 : 2 x + h \to 2 x Answer: f'(x) = 2 x

Example 2 — checking against earlier numbers

Use f'(x) = 2 x to find the gradient of y = x^{2} at x = 3 . 1 f'(3) = 2 \times 3 = 6 This matches the shrinking-h table from two lessons ago, where the secant gradient at x = 3 was heading straight for 6 .

Practice

Using f'(x) = 2 x for f(x) = x^{2}, find f'(4).

Your answer