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Calculus / English

Visualizing the chain rule and product rule | Chapter 4, Essence of calculus

3Blue1Brown · YouTube · 15:56

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Stacked heights, an expanding rectangle and successive number-line mappings explain how local increments lead to sum, product and composition rules. This video visualizes the chain rule of calculus using a three-tiered number line model to track infinitesimal changes. By starting with a concrete value x=1.5x = 1.5 and applying a tiny nudge dxdx, it demonstrates how this change propagates through nested functions like squaring and sine. The animation shows that the total change in the composite function is the product of the local scaling factors at each stage. This leads to the general formula for the derivative of a composition g(h(x))g(h(x)). The video concludes by emphasizing that understanding these mechanical rules requires active practice rather than passive viewing.

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Chapters

0:00Three function combinations1:49Differentiating a sum4:11The rectangle product model5:50The higher-order corner8:18Constant multiples and composition9:06Successive local rates10:00Tracing Infinitesimal Changes Through Nested Functions12:10Deriving the General Chain Rule Formula14:34Summary of Differentiation Rules and Practice Advice

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

Combining simple functions by sums, products and composition builds more complex expressions. The video follows small input changes through these three operations, using geometry to explain differentiation rules. The functions involved must be differentiable at the relevant points.

For a sum, stack the two output heights. A common input increment produces an exact sum of output increments, so (f+g)′=f′+g′(f+g)^{\prime}=f^{\prime}+g^{\prime}. Sine plus a square has derivative cos⁡x+2x\cos x+2x. This is algebraic linearity, not statistical independence.

For a product, imagine a rectangle with sides f(x)f(x) and g(x)g(x). Both sides change with the input. The exact area increment is gΔf+fΔg+ΔfΔgg\Delta f+f\Delta g+\Delta f\Delta g: two strips and a small corner. Multiplying the two derivatives would miss the strips.

Divide by the input increment h. Differentiability makes both Δf and Δg of order h, so their corner product divided by h tends to zero. The surviving terms give f′g+fg′f^{\prime} g+fg^{\prime}. Positive lengths make the picture convenient, while the algebra also works for signed function values.

When one factor is constant its derivative is zero, giving (cf)′=cf′(cf)^{\prime}=cf^{\prime}. Feeding one function’s output into another is a different operation from multiplying their outputs.

Number-line mappings show composition as successive local distance scaling. The overall sensitivity combines both stages. The outer rate must be evaluated at the inner function’s output; the second part of the video develops this chain-rule idea.

Three number lines track x, h=xh=x² and sin h. Near x=1.5x=1.5 the inner rate is 2x=32x=3, while the outer rate is cos⁡(2.25)\cos (2.25), which is negative. A small positive input move increases the square but locally decreases the final sine value. The overall derivative is 2xcos⁡(x2)2x\cos(x^2): evaluate the outer derivative at x², not at x.

If h is differentiable at x and g at h(x)h(x), the composite derivative is g′(h(x))h′(x)g^{\prime}(h(x))h^{\prime}(x). This composes two local linear changes and also holds when the inner derivative is zero. Leibniz notation resembles cancellation, but the justification comes from differentiable increments and limits, not dividing by an ordinary nonzero dh.

Identify the expression’s structure before differentiating: sums use the sum rule, products the product rule, and nesting the chain rule. Complex expressions combine these rules in stages. The closing advice is to practice applying them, connecting the geometric reasoning with calculation.

Knowledge cards

01

The sum rule

Differentiability at the point gives linearity over addition.

(f+g)′=f′+g′(f+g)^{\prime}=f^{\prime}+g^{\prime}
02

Exact product increment

The two strips and the corner give an exact increment, not a product of derivatives.

Δ(fg)=gΔf+fΔg+ΔfΔg\Delta(fg)=g\Delta f+f\Delta g+\Delta f\Delta g
03

The product rule

The corner disappears in the difference-quotient limit, leaving both first-order contributions.

(fg)′=f′g+fg′(fg)^{\prime}=f^{\prime} g+fg^{\prime}
04

Constant multiples

A constant has derivative zero, reducing the product rule to constant scaling.

(cf)′=cf′(cf)^{\prime}=cf^{\prime}
05

Three-Tiered Number Line Model

The input x maps to h=xh=x² and then to sin h. Each local increment is governed by its derivative at the appropriate input.

x→h→g(h)x \to h \to g(h)
06

Local Linear Approximation via Differentials

Differentiability gives a linear leading term plus a smaller remainder. A differential df=f′(a)dx is the linear map itself; the finite increment is approximated by it.

f(a+h)−f(a)=f′(a)h+o(h)f(a+h)-f(a)=f\prime(a)h+o(h)
07

Chain Rule Statement

If u=g(v)u=g(v) and v=h(x)v=h(x) are both differentiable, then the composition F(x)=g(h(x))F(x)=g(h(x)) satisfies F′(x)=g′(h(x))h′(x)F'(x)=g'(h(x))h'(x). Equivalently in differential form, dF=(du/dv)(dv/dx)dxdF = (du/dv)(dv/dx)dx. Crucially, g′g' must be evaluated at the current value of the inner function h(x)h(x), not directly at xx. The proof relies on expressing increments exactly as ΔF=[g′(v)+ϵ]Δv\Delta F = [g'(v)+\epsilon]\Delta v with ϵ→0\epsilon\to0, showing error terms vanish faster than Δx\Delta x, preserving equality in the limit rather than relying solely on informal fraction cancellation.

ddx[g(h(x))]=g′(h(x)) h′(x)\frac{d}{dx}[g(h(x))] = g'(h(x))\,h'(x)
08

Chain notation and its justification

Leibniz notation helps remember the composition of rates, but ordinary fraction cancellation is not its proof. The inner derivative can be zero and the chain rule still holds by differentiability.

dgdx=dgdhdhdx\frac{dg}{dx} = \frac{dg}{dh}\frac{dh}{dx}
09

Differentiation Toolkit Hierarchy

Identify sums, products and composition, then combine rules layer by layer. Keep the outer derivative’s evaluation point and both first-order product terms.

(f+g)′=f′+g′,(fg)′=f′g+fg′,(g∘h)′=(g′∘h)h′(f+g)'=f'+g',\quad(fg)'=f'g+fg',\quad(g\circ h)'=(g'\circ h)h'

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  • Chain rule ExplanationAt 12:25
    Why this connection?

    If u=g(v)u=g(v) and v=h(x)v=h(x) are both differentiable, then the composition F(x)=g(h(x))F(x)=g(h(x)) satisfies F′(x)=g′(h(x))h′(x)F'(x)=g'(h(x))h'(x). Equivalently in differential form, dF=(du/dv)(dv/dx)dxdF = (du/dv)(dv/dx)dx. Crucially, g′g' must be evaluated at the current value of the inner function h(x)h(x), not directly at xx. The proof relies on expressing increments exactly as ΔF=[g′(v)+ϵ]Δv\Delta F = [g'(v)+\epsilon]\Delta v with ϵ→0\epsilon\to0, showing error terms vanish faster than Δx\Delta x, preserving equality in the limit rather than relying solely on informal fraction cancellation.

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