Derivatives and the chain rule

Height and gradient are different quantities

For y = x² at x = 3, the point’s height is 9, while its tangent gradient is 6. The derivative f′(a) gives the local rate of change at input a. The function value f(a) gives the point’s output. A tangent line needs both: it passes through (a,f(a)) with gradient f′(a).

If x measures time in hours and y measures a quantity in units, f(a) has units and f′(a) has units per hour. A rate is not the amount present. A negative derivative means the quantity is decreasing locally; it does not by itself mean the quantity is negative.

Powers include roots and reciprocals

Standard derivatives are (xⁿ)' = nxⁿ⁻¹ on a differentiable real interval, (e^x)' = e^x and (ln x)' = 1/x for x > 0. Constants differentiate to zero; sums, differences and constant multiples work term by term.

Rewrite roots and reciprocals as powers before using the rule. For x > 0:

Multiply by the original exponent, then subtract one from that exponent. Differentiating x⁻² does not just change its sign or leave the exponent unchanged. The reciprocal example also has this derivative for x < 0, away from its excluded input zero.

The power rule applies where the real function is differentiable. Although square root of x has a value at zero, its finite derivative formula does not apply there. The input conditions of the function and the places where a finite derivative exists need separate attention.

Odd roots also have a real meaning at negative inputs. On x < 0, x^(1/3) is the negative real cube root, and its derivative is 1/(3x^(2/3)) > 0. Here x^(2/3) = (cube root of x)² is positive; zero is excluded from this finite derivative formula. In contrast, square root of x² = |x| = -x on x < 0, so its derivative there is −1. Replacing square root of x² by x would discard the input condition and reverse the sign.

For y = 3x²-4 square root of x + e^x on x > 0, the derivative is 6x-2/square root of x + e^x. Keep the original function’s valid inputs when interpreting a result.

A composition multiplies two rates

For y = (x² + 1)³, set u = x² + 1, so y = u³. The outer rate is dy/du = 3u² and the inner rate is du/dx = 2x. Multiplying gives

The outer rate is per unit change in u, whereas we need the change per unit x. The inner rate converts between them. It is not an optional factor appended to a memorised answer.

Similarly (e^(2x-1))' = 2e^(2x-1) and (ln(3x + 4))' = 3/(3x + 4) for x > −4/3. The logarithm condition comes from its original argument 3x + 4 > 0. A derivative expression having a non-zero denominator is not enough to make an original logarithm valid.

For y = e^(-2x), the inner rate is −2, giving y' = -2e^(-2x). The function stays positive but decreases. For y = square root of (5-2x) with x < 5/2, the derivative is -1/square root of (5-2x): the outer factor 1/2 and inner factor −2 combine.

If an expression can be expanded into a polynomial, expand it before using termwise rules. For y = x(x + 2) = x² + 2x, the derivative is 2x + 2. Multiplying the two individual derivatives would give 1, which already fails at x = 0, where the correct gradient is 2. Multiplication of rates in a composition does not authorise multiplying derivatives of two factors.

The inner function can sit inside a root

The same chain rule applies to a rational outer power. Where g(x)^r is real and differentiable,

For f(x) = square root of (x² + 1) = (x² + 1)^(1/2), the inner derivative is 2x. Thus

The expression 1/(2 square root of (x² + 1)) gives the outer rate per unit change in the inner value; it omits the conversion to a rate per unit x. Since x² + 1 > 0 for every real x, this derivative is valid even at x = 0, where it is zero.

For h(x) = (x² + 1)^(3/2), the derivative is 3x square root of (x² + 1). Since x² + 1 = (square root of (x² + 1))², the expression 3x(x² + 1)/square root of (x² + 1) reduces to the same derivative. The denominator never vanishes here, so this rearrangement introduces no missing input.

Root simplification also needs the principal real branch. For example, square root of ((x-1)⁴) = (x-1)² for every real x, because (x-1)² is nonnegative. Its first derivative is 2(x − 1), and its second derivative is 2, including at x = 1. In contrast, square root of ((x-1)²) = |x-1| changes its slope there; it cannot be replaced by x − 1 on an interval that crosses 1.

Differentiate the resulting function again

The second derivative f''(x) is the rate of change of the first derivative. For f(x) = x³, first find f'(x) = 3x², then differentiate that result to get f''(x) = 6x. Differentiating the original function a second time from scratch would only give the first derivative again.

Keep the chain factor on each differentiation. For g(x) = e^3x, g'(x) = 3e^3x and g''(x) = 9e^3x. For h(x) = ln(2x + 1) with x > −1/2, h'(x) = 2(2x + 1)⁻¹ and h''(x) = -4(2x + 1)⁻². The negative sign comes from the new exponent −1; the second factor 2 comes from the inner rate. These examples need only the power rule, constant multiples and the chain rule.

An equivalent formula must cover the stated inputs

Simplification can hide a missing input. A proposed expression must be defined at every input where the requested derivative is required, as well as have the correct value there.

Form the tangent from a point and gradient

For y = x² − 3x + 1 at x = 2, the point is (2,−1), and y′ = 2x − 3 gives gradient 1. The tangent is y + 1 = x − 2, or y = x − 3. Substitute x = 2 to check the point, then check the line’s coefficient agrees with the derivative.

In general, at a differentiable point with input a,

The variable x in this line is free to vary; a is the fixed input where the tangent touches the curve. A line passing through the point is not necessarily the tangent: its gradient must agree too. Conversely, y = f′(a)x usually misses the point because it silently assumes a zero intercept.

Use a numerical derivative at the requested input

A graphing calculator or graphing software can approximate the derivative without first displaying an algebraic formula. Enter the whole original expression, select the derivative-at-a-point operation, and supply the input explicitly. A cursor’s current graph height is not its tangent gradient.

For f(x) = ln(3x + 4) at x = 2, the function value is ln 10 ≈ 2.302585, whereas the derivative is 3/10 = 0.3. A numerical derivative near 0.3 agrees with the analytic calculation. The tangent is y- ln 10 = 0.3(x-2), or y = 0.3x + ln 10-0.6.

Record the function, input and requested precision. Keep the exact logarithm in an exact tangent equation; a rounded derivative or function value produces only an approximate line. If a numerical operation fails at an excluded input, increasing display precision does not make that original function valid there.

Interpret a derivative sign locally

A positive derivative throughout an interval shows the function increasing there; a negative derivative shows it decreasing. A zero derivative at a point identifies a stationary candidate, not automatically a maximum or minimum.

These statements describe the original function’s rise, fall or horizontal tangent. They do not say that a positive function value forces a positive derivative. The positive decreasing exponential above supplies a counterexample.