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.

At SL, estimate $\lim_{x\to a}f(x)$ from values approaching $a$ on both sides or from the graph; formal analytic limit methods are not required. The derivative is the limiting gradient and may be written $dy/dx$, $f'(x)$, $dV/dr$ or $ds/dt$, with units of output per unit input. A limit can exist even when the displayed point is missing, because it describes nearby behaviour rather than only $f(a)$.

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 $d(ax^n)/dx=anx^{n-1}$ term by term for integer exponents in this objective, retaining coefficients and differentiating constants to zero.

The derivative of $x^0$ is zero. Rational powers and chain-rule factors belong to SL 5.6, so do not import them into this integer-power step.

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.

A boundary condition selects the constant: if $dy/dx=3x^2+x$ and $y(1)=10$, then $y=x^3+x^2/2+C$ gives $C=8.5$. A definite integral accumulates signed change; when $f(x)>0$, technology can evaluate $\int_a^b f(x)\,dx$ as the area between the curve and the $x$-axis. Write the correct integral before calculating.

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.

Core derivatives: $d(x^n)/dx=nx^{n-1}$ for rational $n$ where defined, $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(e^x)'=e^x$, and $(\ln x)'=1/x$. Also $(u/v)'=(u'v-uv')/v^2$. Example: for $y=x^2e^x$, the product rule gives $y'=e^x(x^2+2x)$; for $y=\ln(3x+1)$, the chain rule gives $3/(3x+1)$.

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.

Classification: if $f'$ changes from positive to negative, the stationary point is a local maximum; negative to positive gives a local minimum. Alternatively, at $f'(a)=0$, $f''(a)>0$ implies a local minimum and $f''(a)<0$ a local maximum. A point of inflexion requires a change in concavity, so $f''=0$ alone is not sufficient; for example $y=x^4$ has $f''(0)=0$ but no concavity change. Use 'concave-up' for $f''>0$ and 'concave-down' for $f''<0$.

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.

From $t_1$ to $t_2$, displacement is $\int_{t_1}^{t_2}v(t)\,dt$, while total distance is $\int_{t_1}^{t_2}|v(t)|\,dt$. Split the distance integral wherever $v(t)=0$. Speed is $|v|$: an object slows when velocity and acceleration have opposite signs and speeds up when their signs agree.

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.

Formula set: $\int x^n\,dx=x^{n+1}/(n+1)+C$ for $n\ne-1$, $\int dx/x=\ln|x|+C$, $\int e^x\,dx=e^x+C$, $\int\sin x\,dx=-\cos x+C$, and $\int\cos x\,dx=\sin x+C$. For a linear composite, account for the inner gradient: $\int\cos(2x+3)\,dx=\tfrac12\sin(2x+3)+C$. Reverse-chain recognition gives $\int k g'(x)[f(g(x))]\,dx$ by substitution or inspection.

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.

Fundamental evaluation: if $F'=f$, then $\int_a^b f(x)\,dx=F(b)-F(a)$. For area between $y=f(x)$ and $y=g(x)$, first find intersections, then integrate top minus bottom on each interval: $A=\int_a^b|f(x)-g(x)|\,dx$, splitting wherever their order changes. This preserves geometric area even when the signed integral cancels.