A derivative measures local change
A straight line has one constant , but a curve's steepness can change from point to point. The of the secant through nearby graph points is a change in y divided by a change in x. The derivative describes the limiting value of that ratio as the separation tends to zero. For, write f′(x) or dy/dx. It is a function giving a local , not merely the original y-value at the point.
A limit describes what a function approaches as its input approaches a value. For, the expression is undefined at x = 1, but for x ≠ 1 it equals x + 1. Hence its limit as x approaches 1 is 2. A limit need not equal a function's value at the point, and the function need not be defined there. For a polynomial, direct substitution finds a finite limit because polynomials are continuous everywhere.
Derive a simple rule
For, the over a small increment h is [(x + h)² − x²]/h. Expanding gives for h ≠ 0. As h tends to zero, this approaches 2x. Thus d(x²)/dx = 2x. The method does not set h = 0 before cancellation, which would create division by zero. It takes the limit of a ratio defined for nonzero h.
The rule is wherever the expression is differentiable. Thus, d(x)/dx = 1 and d(7)/dx = 0. A constant has zero derivative because its graph has no vertical change. Constant factors carry through, and derivatives of sums are found term by term. Therefore the derivative of 3x⁴ − 5x² + 7x − 9 is 12x³ − 10x + 7. Differentiation does not multiplying every by the same number.
Handle roots, reciprocals and brackets
Rewrite roots and reciprocals using powers before applying the rule. Since, its derivative is (1/2)x⁻¹ᐟ² = 1/(2√x) for x > 0. Since, its derivative is −x⁻² for x ≠ 0. The domain restrictions remain important: a derivative formula is not valid where its original function or the derivative itself is undefined. In elementary exercises, clearly state a positive-root domain when required.
The chain rule differentiates a composite: if and, then dy/dx = (dy/du)(du/dx). For, differentiate the outer cube and multiply by the inner derivative 2, giving 6(2x + 1)². Expanding first would give the same answer but take longer. A frequent error is to write 3(2x + 1)² and forget the inner factor. Check a small example against expansion until the rule feels justified.
Products, quotients and trigonometric functions
For differentiable u and v. It is not u′v′. For a quotient with v ≠ 0. With, the derivative is [(x + 1) − x]/(x + 1)² = 1/(x + 1)², excluding x = −1. Set out the numerator in brackets before simplifying, because the minus sign applies to the whole uv′ term.
With angles in radians, and d(tan x)/dx = sec²x where defined. The radian condition is essential. For, the chain rule gives 3 cos 3x, with x in radians. These rules do not transfer unchanged to a function whose numerical argument is measured in degrees. If an examination problem supplies a different convention or restricts the required functions, follow that explicit instruction.
Tangents, normals and rates
To find a tangent at, first find the point (a, f(a)), then calculate f′(a), then use y − f(a) = f′(a)(x − a). For at x = 2, the point is (2, 4) and 4, so the tangent is. A nonvertical normal has the negative reciprocal of the tangent . If the tangent is horizontal, the normal is vertical, so use rather than dividing by zero.
A derivative can represent a rate with units. If s(t) is in metres and time in seconds, is ds/dt in m/s, and is d²s/dt² in m/s². For, is 3t² − 3 and 6t. may be negative because it records direction; is its magnitude. A stationary object at an instant has zero , but its need not be zero and its need not vanish.
Classify stationary points and optimise
A stationary point satisfies f′(x) = 0. To classify it, inspect how f′ changes sign. Negative to positive gives a local minimum; positive to negative gives a local maximum. A second-derivative test is convenient: f″ > 0 at a stationary point indicates a local minimum and f″ < 0 a local maximum. If f″ = 0, that test is inconclusive. For, x = 0 is stationary but is not a maximum or minimum.
For a rectangle of perimeter 20 m, let sides be x and 10 − x, with 0 < x < 10. Area gives A′ = 10 − 2x, so the stationary point is x = 5. Since A″ = −2, the area is maximised, giving a square of area 25 m². In an optimisation problem, state the physical domain and inspect endpoints if included. Self-check: differentiate 4x³ − 2x + 6 and find the minimum point of. Answers are 12x² − 2 and (3, −1). The final coordinates require substituting the stationary x-value back into the original function.
Use sign changes when the second derivative is zero
For, and. At x = 0 both derivatives vanish, so the second-derivative test is inconclusive. However, f′ is negative to the left and positive to the right, establishing a minimum at (0, 0). For, f′ = 3x² is positive on both sides of zero, so its stationary point is not an extremum. “Second derivative zero” is a request for another test, not a classification.
On a closed interval, a largest or smallest value may occur at an endpoint rather than a stationary point. Evaluate the original function at every candidate and permitted endpoint, then compare values. For a function increasing throughout [1, 3], the minimum is at 1 and maximum at 3 even though no derivative zero occurs inside the interval.