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Pure · Differentiation

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The chain rule

Differentiating a function of a function — y = f(g(x)) — by multiplying the outside derivative by the inside derivative, with the dy/dx = (dy/du)(du/dx) substitution method and the reciprocal connection dx/dy = 1 / (dy/dx).

Pure · Differentiation

The chain rule

Differentiating a function of a function — y = f(g(x)) — by multiplying the outside derivative by the inside derivative, with the dy/dx = (dy/du)(du/dx) substitution method and the reciprocal connection dx/dy = 1 / (dy/dx).

Why it works

The power rule handles xnx^n, but real functions are usually nested: a curve like y=(3x+1)5y = (3x + 1)^5 is one function — "raise to the fifth" — wrapped around another — "triple and add one". You could expand the bracket and differentiate term by term, but for (3x+1)5(3x+1)^5 that is grim, and for y=x2+1y = \sqrt{x^2 + 1} it is impossible. The chain rule differentiates the nesting directly.

Give the inside its own name. Write u=3x+1u = 3x + 1, so the outer function is y=u5y = u^5. Now you have two easy derivatives: dydu=5u4,dudx=3.\frac{dy}{du} = 5u^4, \qquad \frac{du}{dx} = 3. The chain rule says multiply them: dydx=dydu×dudx=5u4×3=15(3x+1)4.\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx} = 5u^4 \times 3 = 15(3x+1)^4.

Why does multiplying rates work? Think of the derivatives as conversion factors. If yy changes 5u45u^4 times as fast as uu, and uu changes 33 times as fast as xx, then yy changes 5u4×35u^4 \times 3 times as fast as xx — the rates compound, exactly the way "miles per hour" times "hours per day" gives "miles per day". In Leibniz notation the dudu's look like they simply cancel, and that picture is a faithful guide even though dydx\frac{dy}{dx} is not really a fraction.

In practice you rarely write uu out. The working version is: differentiate the outside (leaving the inside alone), then multiply by the derivative of the inside. For y=(3x+1)5y = (3x+1)^5: the outside gives 5(3x+1)45(3x+1)^4, the inside 3x+13x+1 differentiates to 33, so dydx=15(3x+1)4\frac{dy}{dx} = 15(3x+1)^4. The general pattern for a bracket to a power is worth memorising: ddx[f(x)]n=n[f(x)]n−1 f′(x).\frac{d}{dx}\big[f(x)\big]^n = n\big[f(x)\big]^{n-1}\,f'(x).

The reciprocal connection. Sometimes xx is given as a function of yy and you want dydx\frac{dy}{dx} without rearranging. Because the rates are reciprocals, dydx=1 dxdy .\frac{dy}{dx} = \frac{1}{\,\frac{dx}{dy}\,}. If x=y3+2yx = y^3 + 2y, then dxdy=3y2+2\frac{dx}{dy} = 3y^2 + 2, so dydx=13y2+2\frac{dy}{dx} = \dfrac{1}{3y^2 + 2} straight away. This is the same idea as the chain rule with the roles of xx and yy swapped, and it is the key to differentiating inverse functions.