The chain rule

Drag the point on the top curve. Whatever value it lands on, $u$, feeds straight into the bottom curve — the point down there is never dragged directly, it's just wherever $f$ sent it.

Call the whole composition $h(x) = g(f(x))$. Its derivative is the chain rule —

• $h(x)$ — the whole composition, $h(x) = g(f(x))$.
• $u$ — the inner value, $u = f(x)$, that feeds into $g$.
• $\frac{dg}{du}$ — the outer function's slope, evaluated at the inner value $u$.
• $\frac{du}{dx}$ — the inner function's slope with respect to $x$.

1. Multiply the two local slopes

   In words: the combined rate of change is the outer function's slope (at the inner value $u$) multiplied by the inner function's slope. Two local slopes, multiplied, give the slope of the whole chain.

Drag the top point and watch both slopes update — then check that their product really does track $dh/dx$ as you move.

Here $f(x) = 2x + 1$ and $g(u) = u^2$, so $h(x) = (2x+1)^2$.

1. Find the inner value at x = 1

   $$u = f(1) = 2(1) + 1 = 3$$
2. Find both local slopes

   $$g'(u) = 2u = 6, \qquad f'(x) = 2$$
3. Multiply them (the chain rule)

   $$\frac{dh}{dx} = 6 \times 2 = 12$$
4. Check directly, a different route

   Expand the square first: $(2x+1)^2 = (2x+1)(2x+1) = 4x^2+2x+2x+1 = 4x^2+4x+1$.

   Differentiate term by term:

   • $\frac{d}{dx}[4x^2]=8x$
   • $\frac{d}{dx}[4x]=4$
   • $\frac{d}{dx}[1]=0$

   Adding those up: $h'(x)=8x+4$.

   Evaluating at $x=1$: $h'(1) = 8(1)+4 = 12$ — same answer, two routes.