[frac d/dx] e[sup x] = e[sup x], [frac d/dx] ln x = [frac 1/x], [frac d/dx] sin x = cos x, [frac d/dx] cos x = −sin x, [frac d/dx] tan x = sec[sup 2] x. Angles are always in radians in calculus.

With the chain rule: ddx e^f(x) = f′(x) e^f(x), ddx ln f(x) = f′(x)f(x), ddx sin f(x) = f′(x) cos f(x) and ddx cos f(x) = −f′(x) sin f(x).

Worked example. y = e^3x². Here f(x) = 3x², so dydx = 6x e[sup 3x²]. Similarly ddx ln(x² + 1) = 2xx² + 1.

Product rule: (uv)′ = u′v + uv′. Quotient rule: (uv)′ = u′v − uv′v². Chain rule: dydx = dydu × dudx.

Decide the outer structure first (a product, a quotient or a function of a function), then differentiate each piece with the standard results.

Worked example. y = x² e^−x. Product rule with u = x², v = e^−x: dydx = 2x e^−x − x² e^−x = x e[sup −x](2 − x).

The gradient of the tangent at x = a is f′(a); the tangent is y − f(a) = f′(a)(x − a). Stationary points occur where f′(x) = 0; classify them with the second derivative (f″ < 0 maximum, f″ > 0 minimum) or a table of gradients.

If Q depends on time t, then dQdt is the rate of change of Q, with units of Q per unit of time.

Worked example. Stationary points of y = e^x − 2x: y′ = e^x − 2 = 0 gives x = ln 2, y = 2 − 2 ln 2. Since y″ = e^x > 0, it is a minimum turning point.