Trigonometric and exponential calculus

Standard derivatives have conditions

The power, chain, product and quotient rules still apply when a function contains trigonometric, exponential or logarithmic terms. First identify the original domain and the changing input, then choose the rule. The following trigonometric derivatives use radians.

The derivative of tan x is undefined at the same excluded inputs as the original function: x = π/2 + kπ for integer k. A formula’s appearance does not remove an original restriction.

Do not change angle mode silently

If θ is a numerical angle measured in degrees, sin(θ°) means sin(πθ/180) when expressed in radians. The chain rule therefore gives

At θ = 0, the derivative with respect to that degree-valued input is π/180, rather than 1. Changing a calculator’s display setting does not change the variable in a mathematical model. Unless stated otherwise below, trigonometric inputs are in radians.

An outer derivative needs the inner rate

For y = sin(3x + 1), the chain rule gives y′ = 3 cos(3x + 1). The 3 is the rate at which the inner angle changes with x. Similarly, y = e^(2x-5) gives y' = 2e^(2x-5), and y = tan(2x) gives y′ = 2sec²(2x) wherever cos(2x) ≠ 0.

For y = ln(4x + 3), the original logarithm requires 4x + 3 > 0, hence x > −3/4. Its derivative is 4/(4x + 3) on that domain. The rational expression can be evaluated for some other x-values, but it cannot supply derivatives at points where the original logarithm does not exist.

The input to the logarithm must be positive; x itself need not be. For y = ln(x²), the original domain is x ≠ 0 and

This derivative is valid on either side of zero. Replacing ln(x²) by 2ln x would incorrectly discard the negative branch; the identity valid on the full original domain is ln(x²) = 2ln|x|.

Combine the standard and algebraic rules

For y = x sin x, both factors change. The product rule gives y′ = sin x + x cos x, rather than x cos x alone. For y = x ln x, it gives y′ = ln x + 1 on x > 0.

For y = (ln x)/x, use the quotient rule with numerator derivative 1/x and denominator derivative 1:

The denominator condition x ≠ 0 alone would be insufficient: the original numerator still requires x > 0. A simplified answer keeps the full original domain.

A derivative also supports decisions about the function. For y = sin(2x) on 0 ≤ x ≤ π/2, y′ = 2cos(2x). Its only interior zero is x = π/4. The derivative changes from positive to negative there, so y = 1 is the maximum; both endpoint values are zero. This agrees with the known range of sine and uses the stated radian interval.

Reverse the inner scale when integrating

An antiderivative G of f satisfies G′ = f. Since adding a constant does not change a derivative, the indefinite integral represents G + C. Use the derivative relationships above in reverse, and verify the proposed result by differentiating it.

For a linear input ax + b with a ≠ 0, the constant inner derivative must be undone:

For example, ∫cos(3x + 1) dx = sin(3x + 1)/3 + C. Differentiation multiplies by 3 and recovers the original cosine. For ∫ e^(-2x) dx, the result is -e^(-2x)/2 + C: differentiating multiplies by −2, so the two negative signs cancel. If a = 0, integrate the constant integrand directly rather than dividing by zero.

Also ∫sec²x dx = tan x + C on a continuous tangent branch. For example, on −π/2 < x < π/2 the derivative of tan x is the required sec²x. This formula cannot justify an ordinary definite integral across x = π/2, where the integrand is undefined, merely because tangent values can be written at the two endpoints.

The missing n = −1 power case is

on an interval not crossing zero. For x > 0, this is ln x + C. For x < 0, ln|x| = ln(−x), whose derivative is (1/(−x))(−1) = 1/x. The two intervals may have different constants; neither expression defines a value at zero.

A condition selects a member of a family. For example, y′ = eˣ gives y = eˣ + C. If y(0) = 3, then 3 = 1 + C, so y = eˣ + 2. Check both the derivative and the given point.