5.1 Calculus - SL content

A derivative is a local rate of change.

The derivative is the limit of average change as the interval shrinks; it describes the tangent slope at a point.

Example

For f(x)=x², (f(1+h)−f(1))/h=2+h, so f′(1)=2 as h→0.

State the variable and point before differentiating, then interpret the sign and units of the result.

A derivative at one point is not the total change across an interval.

The first derivative reveals where a function rises or falls.

If f′(x)>0 the graph increases and if f′(x)<0 it decreases; sign changes identify possible stationary extrema.

Example

For f′(x)=x−2, the function decreases before x=2 and increases after it, so x=2 is a local minimum.

Make a sign chart around every critical value rather than relying on the derivative being zero alone.

A stationary point can be a saddle or inflection; f′=0 is a candidate, not a classification.

The power rule differentiates xⁿ by lowering the exponent.

For a constant n, d(xⁿ)/dx=nxⁿ⁻¹; the factor n records how rapidly the power changes.

Example

d(3x⁵−2x²)/dx=15x⁴−4x.

Apply the rule term by term and keep constants; check the domain of any fractional or negative power.

The derivative of x⁰ is zero, and the rule does not remove a chain-rule factor from a composite function.

A tangent uses the derivative; a normal uses its negative reciprocal.

At x=a, the tangent slope is f′(a); a perpendicular normal has slope −1/f′(a) when the tangent is neither horizontal nor vertical.

Example

For y=x² at x=1, the tangent is y−1=2(x−1), while the normal is y−1=−(x−1)/2.

Find the point and tangent slope first, then use point–slope form for the requested line.

A vertical tangent has a horizontal normal, so the reciprocal-slope formula needs a separate case.

Integration reverses differentiation up to a constant.

An antiderivative F satisfies F′=f; all antiderivatives differ by a constant because differentiation removes constants.

Example

∫(6x²−4)dx=2x³−4x+C; differentiating checks the result.

Integrate each term, include +C for an indefinite integral and use a condition to determine C.

An indefinite integral is a family, not one function until a condition is supplied.

Product, quotient and chain rules preserve how functions are built.

Differentiate the outer and inner structure rather than expanding blindly: (uv)′=u′v+uv′ and (f(g(x)))′=f′(g)g′.

Example

For y=(x²+1)³, y′=3(x²+1)²·2x=6x(x²+1)².

Name the outer operation and inner function before applying a rule.

The derivative of a product is not the product of derivatives.

The second derivative describes curvature and acceleration.

f″ measures how the slope changes; its sign indicates concavity and, in kinematics, acceleration.

Example

For f(x)=x³, f″=6x, so the graph is concave down for x<0 and up for x>0.

Use f′ for increasing/decreasing and f″ for curvature; test continuity when claiming an inflection.

A positive second derivative does not mean the function itself is positive.

Optimization compares critical points with the feasible boundary.

A maximum or minimum occurs at a critical point or endpoint of the allowed domain; the model and constraints decide which is meaningful.

Example

For a rectangle with perimeter 20, A=x(10−x) is largest at x=5, not at an unconstrained value outside 0≤x≤10.

Differentiate, solve candidates, then compare all endpoints and check units.

A local maximum need not be the global maximum when the domain is restricted.

Kinematics links position, velocity and acceleration by differentiation.

If s(t) is position, v=s′ and a=v′=s″; signs depend on the chosen positive direction.

Example

For s=4t²−t, v=8t−1 and a=8, so velocity changes while acceleration stays constant.

State the time interval and direction convention before interpreting motion.

Negative velocity means motion opposite the chosen direction, not necessarily slowing down.

Indefinite integration reconstructs a family of functions.

Reverse the derivative term by term and retain the arbitrary constant; the constant represents the unknown vertical shift.

Example

∫(2x+sin x)dx=x²−cos x+C.

Differentiate the answer and use any initial value to fix C.

Do not omit C when no initial condition has selected one antiderivative.

A definite integral accumulates signed change over an interval.

∫ₐᵇf(x)dx is net signed area; geometric area may require splitting where f changes sign.

Example

∫₋¹¹x dx=0 by cancellation, although the total geometric area is 1.

Find roots and compare the requested quantity—net area, total area or accumulated change.

A negative integral is possible; it does not mean an area has become physically negative.