A hypothetical square has side u(t)=2t+1, growing by 2 cm each second. Why is its area A=u² increasing by 12 cm²/s at t=1? Change passes through two dependencies: time changes the side, which changes area. Both rates matter. t counts seconds and u centimetres, t≥0. Prerequisites: exponentials, logarithms, trigonometric identities and derivatives.

Combining changes

Let f and g be differentiable at the required point and c constant. A sum combines rates; a product changes through both factors:

(f ± g)′ = f′ ± g′, (cf)′ = cf′; (fg)′ = f′g + fg′.

A quotient needs a nonzero denominator:

((f)/(g))′ = (f′g − fg′)/((g)^(2)) (g ≠ 0).

These are not f′g′ or f′/g′: even x·x would incorrectly give 1 instead of 2x. Rules and derivations.

Multiply outer and inner rates

For A(u(t)), evaluate the outer derivative at u(t), then multiply by u′(t):

(dA)/(dt) = (dA)/(du)(du)/(dt) = 2u ⋅ 2 = 4(2t + 1).

At t=1, u=3, dA/du=6 cm²/cm and du/dt=2 cm/s, giving 12 cm²/s. The notation resembles cancelling units, but this is a theorem for differentiable functions, not permission to cancel arbitrary symbols. OpenStax

Two successive rates

t, su=2t+1, cmA=u², cm²du/dt=2 cm/sdA/du=2u cm
At t=1: u=3 cm, dA/du=6 cm²/cm; 6·2=12 cm²/s. Dependency diagram, not to scale; arrows do not indicate motion.

A table with conditions

For real r the following power rule is safe for x>0. Positive integer powers also have their polynomial derivative at negative x and zero; differentiate a constant separately.

C′ = 0, ((x)^(r))′ = r(x)^(r − 1) (x > 0).

Trigonometric arguments are radians. The relevant denominator for tan/cot must be nonzero:

(sin⁡x)′ = cos⁡x, (cos⁡x)′ = − sin⁡x.

(tan⁡x)′ = (1)/((cos⁡)^(2)x), (cot⁡x)′ = − (1)/((sin⁡)^(2)x).

For exponentials and logarithms b>0, b≠1; logarithms also require x>0:

((e)^(x))′ = (e)^(x), ((b)^(x))′ = (b)^(x)ln⁡b.

(ln⁡x)′ = (1)/(x), ((log⁡)_(b)x)′ = (1)/(xln⁡b).

Table sources: trigonometric derivatives, exponentials and logarithms.

Start by identifying the structure

(xsin⁡x)′ = sin⁡x + xcos⁡x; ((x)/(x + 1))′ = (1)/((x + 1)^(2)).

The second formula retains x≠−1 from the original expression. Another chain:

(ln⁡(2t + 1))′ = (2)/(2t + 1) (t > − 1/2).

Dropping the factor 2 loses the inner rate. Identify the main operation — sum, product, quotient or composition — then differentiate its parts and check domain and units. A table alone does not explain the rate of change.