The Chain Rule
Differentiate a composition by multiplying the rate of the outer function by the rate of the inner one. The rule is what makes substitution, related rates and implicit differentiation possible.
What a learner can do afterwards
- Identify composition layers and apply or check their derivative factors
- Differentiate nested compositions, including exponential and logarithmic functions
- Relate changing quantities through rates and units, including solving for an unknown rate
1 · Read
If g is differentiable at x and f is differentiable at g(x), the derivative of f(g(x)) is f'(g(x))g'(x). Keep the inner input while differentiating the outer function, then multiply by the inner derivative. For (2x²+3)⁴, the factors are 4(2x²+3)³ and 4x, giving 16x(2x²+3)³.
Evaluate only after differentiating. For F(x) = (x²+2)², the derivative is 2(x²+2) times 2x = 4x(x²+2). At 2, it equals 8 times 6 = 48. The power rule lowers the outer exponent; the inner derivative supplies a separate multiplier.
Using radians, the derivatives of sin u and cos u are cos u and -sin u. Thus cos(g(x)) has derivative -sin(g(x))g'(x). The argument remains g(x), not g'(x). Check the argument and the multiplier separately. Differentiating the resulting derivative again gives a second derivative, not the original function.
The elementary rules are (e^u)' = e^u and (ln u)' = 1/u for u > 0. A logarithmic composition needs a positive argument and its inner derivative as a multiplier. For x > 0, cos(ln(3x)) has derivative -sin(ln(3x)) times (1/(3x)) times 3, giving -sin(ln(3x))/x.
For f(g(h(x))), apply the chain rule to f and g(h(x)), then to g and h. The result is f'(g(h(x)))g'(h(x))h'(x), provided the needed derivatives exist. This is two chain-rule applications. If an inner derivative is zero and the other factors are defined and finite, the composite derivative is zero.
If y = f(x) and x changes with time, dy/dt = (dy/dx)(dx/dt). Units cancel like fractions. Metres per metre times metres per second gives metres per second. For rates 5 and -2 in those units, the output rate is -10 metres per second. The sign records the direction of change.
An exponential outer layer retains its entire argument. For e^(cos(3x)), the derivative factors are e^(cos(3x)), -sin(3x) and 3. Multiply them before evaluating an input. At a point, use the values of every factor; a zero inner factor makes the product zero only when the remaining factors are defined and finite.
For a sphere, V = (4/3)pi r³, so dV/dt = 4pi r² dr/dt. Differentiate while r still varies with time. At r = 2 cm and dr/dt = 3 cm/s, dV/dt = 48pi cm³/s. The same equation can be solved for dr/dt when the volume rate and positive radius are known.
For a circle, A = pi r², so dA/dt = 2pi r dr/dt. Form the rate equation before substituting a moment's radius. A radius in metres and radius rate in metres per second give area rate in square metres per second. To infer an inner rate's units, cancel the intermediate quantity from the desired output rate.
Keep the inner arguments and multiply the derivative factors. For changing quantities, differentiate before inserting momentary values.
Original chain-rule lesson and questions written for this learning dataset, with dataset flashcards.
Practice questions written for this learning dataset. The rewritten readings were written for this dataset.
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Where it sits
18 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.