Why does the chain rule multiply?

1. 1. Squeeze it sideways

   The dashed wave is $y = \sin u$. The solid one is $y = \sin(kx)$, for now with $k = 1$: the same wave. $P$ is at $x$ on the solid wave, and its partner $Q$, at $u = kx$ on the dashed one, is at the same height.

   Question: Squeeze the solid wave sideways by $3$: $y = \sin 3x$. $P$ and $Q$ keep the same height. How will the slope at $P$ compare with the slope at $Q$?

   Three times as steep. Near $P$ the solid wave is the dashed wave near $Q$ squeezed into a third of the width: the same rise over a third of the run. Every slope of $\sin 3x$ is $3$ times the slope of $\sin u$ at the partner, $u = 3x$.

   $$\frac{d}{dx}\,f(kx) = k\,f'(kx)$$$\frac{d}{dx}f(kx)$ $=k{f}^{\prime }(kx)$

   The figure: The dashed wave y = sin u and the solid wave y = sin(g(x)) with u = 3x. P at x = 0.40 and its partner Q at u = 1.20 share the height 0.93; the inside's slope there is 3.00. The tangent at P has slope 1.09, the slope at Q, 0.36, times 3.00.

2. 2. The slip: no squeeze

   A common slip differentiates the outside and stops: $\cos 3x$, with no $3$. That copies $Q$'s slope to $P$ unsqueezed, the amber line. At $x = 0.4$, $Q$ is at $u = 1.2$, where the dashed wave's slope is $\cos 1.2 \approx 0.36$.

   Question: What is the slope of $y = \sin 3x$ at $P$, where $x = 0.4$?

   $3\cos 1.2 \approx 1.09$, three times the amber line's slope, and the amber line cuts across the curve. After differentiating the outside, multiply by the inside's derivative, every time: here the $3$ from $u = 3x$.

   The figure: The dashed wave y = sin u and the solid wave y = sin(g(x)) with u = 3x. P at x = 0.40 and its partner Q at u = 1.20 share the height 0.93; the inside's slope there is 3.00. The tangent at P has slope 1.09, the slope at Q, 0.36, times 3.00. In amber, a line at P with Q's slope and no squeeze, cutting across the curve.

3. 3. A curved inside

   Now the inside curves: $u = x^2/2$. Near any point it is almost a straight line of slope $x$, so near $P$ the wave is $\sin u$ squeezed by $x$: the dotted wave, squeezed by exactly that, hugs the curve at $P$. $Q$ is still the point of $\sin u$ at $P$'s height, at $u = x^2/2$.

   Drag $P$ to the first point right of $x = 0.5$ where the curve is flat.

   Flat at $x \approx 1.77$, where $u = x^2/2 \approx 1.57$, which is $\pi/2$: $Q$ is at the peak of $\sin u$. The squeeze there is $1.77$, but a flat slope squeezed is still flat. Read at $x$ instead, the outside's slope $\cos 1.77 \approx -0.20$ is not flat at all.

   The figure: The dashed wave y = sin u and the solid wave y = sin(g(x)) with u = x²/2. P at x = 1.77 and its partner Q at u = 1.57 share the height 1.00; the inside's slope there is 1.77. The tangent at P has slope 0.01, the slope at Q, 0.00, times 1.77.

4. 4. Read the outside at Q

   A second slip reads the outside's slope at $x$ instead of at $Q$. At $x = 3$ it takes the slope of $\sin u$ at $u = 3$, the amber point, times the squeeze, $3$: the amber line at $P$. But $Q$, at $P$'s height, is at $u = 3^2/2 = 4.5$.

   Question: What is the slope of $y = \sin(x^2/2)$ at $x = 3$?

   $3\cos 4.5 \approx -0.63$. Near $P$ the curve is the wave near $Q$ squeezed by $3$, so the outside's slope belongs at $Q$. The amber point at $u = 3$ is at another height, $\sin 3 \approx 0.14$ against $\sin 4.5 \approx -0.98$, with another slope.

   $$\frac{d}{dx}\,f\bigl(g(x)\bigr) = f'\bigl(g(x)\bigr)\,g'(x)$$$\frac{d}{dx}f(g(x))$ $=f^{\prime }(g(x))g^{\prime }(x)$

   The figure: The dashed wave y = sin u and the solid wave y = sin(g(x)) with u = x²/2. P at x = 3.00 and its partner Q at u = 4.50 share the height −0.98; the inside's slope there is 3.00. The tangent at P has slope −0.63, the slope at Q, −0.21, times 3.00. In amber, the point of the dashed wave at u = 3.00, at height 0.14, and the line at P its slope times the squeeze gives, of slope −2.97.

The film, as said, 1:42

1. 0:00

   The dashed wave is $y = \sin u$. The solid one is $y = \sin(kx)$, for now with $k = 1$: the same wave.

2. 0:06

   Squeeze the solid wave sideways by $3$. $P$ stays at $x = 0.4$; its partner $Q$, at the same height, moves out to $u = 3x$.

3. 0:16

   Near $P$, the solid wave is the dashed wave near $Q$ squeezed into a third of the width: the same rise over a third of the run.

4. 0:25

   Run $P$ along: wherever it goes, its slope is $3$ times $Q$'s. So $(\sin 3x)' = 3\cos 3x$.

5. 0:33

   A common slip stops at $\cos 3x$, with no $3$. That copies $Q$'s slope to $P$ unsqueezed: the amber line.

6. 0:42

   At $x = 0.4$, $Q$'s slope is $\cos 1.2 \approx 0.36$. The true slope at $P$ is $3$ times that, about $1.09$.

7. 0:47

   Now let the inside curve: $u = x^2/2$. Its slope is $x$, so the squeeze grows as $P$ moves right.

8. 0:56

   Near $P$ the wave is $\sin u$ squeezed by $x$. The dotted wave, squeezed by exactly that, hugs the curve.

9. 1:03

   Run $P$ to $x \approx 1.77$, where the curve is flat. There $u = x^2/2 \approx 1.57$, which is $\pi/2$: $Q$ is at the peak.

10. 1:11

    A flat slope squeezed is still flat. Read at $x$ instead, the outside's slope, $\cos 1.77 \approx -0.20$, is not flat at all.

11. 1:18

    Run on to $x = 3$, where the squeeze is $3$ and $Q$ is at $u = 4.5$.

12. 1:25

    A second slip reads the outside's slope at $u = 3$, the amber point, instead of at $Q$: $3\cos 3 \approx -2.97$, the amber line.

13. 1:32

    But $Q$, at $P$'s height, is at $u = 4.5$, so the slope is $3\cos 4.5 \approx -0.63$.

14. 1:36

    So the chain rule: the outside's slope, read at the partner $g(x)$, times the squeeze $g'(x)$. That is $f'(g(x))\,g'(x)$.