Composition with an inner product rule

When a product appears inside a composition, one combines the circle diagram and the product rule: one applies the diagram for the outer link, then goes down to differentiate the argument with the ordinary rules. Example: $D[\cos(\sin(e^x-3)\cdot x^2)]$.

Example — Composite function with an inner product rule

Step 1: the outer function is $\cos\triangle$, with $\triangle = \sin(e^x-3)\cdot x^2$.

Outer diagram:

Outer diagram: the right-hand link is $\cos\triangle$, whose derivative is $-\sin\triangle$.

Example (continued)

Step 2: we need $D[\sin(e^x-3)\cdot x^2]$ with the product rule. Setting $f = \sin(e^x-3)$, $g = x^2$: $f'\cdot g + f\cdot g' = D[\sin(e^x-3)]\cdot x^2 + \sin(e^x-3)\cdot 2x.$

Step 3: for $D[\sin(e^x-3)]$ we need another circle diagram:

Inner diagram: $D[\sin(e^x-3)] = e^x\cdot\cos(e^x-3)$.

Example — Final result

$D[\cos(\sin(e^x-3)\cdot x^2)] = -\sin\bigl(\sin(e^x-3)\cdot x^2\bigr)\cdot\bigl[e^x\cos(e^x-3)\cdot x^2 + 2x\sin(e^x-3)\bigr].$

Beware — If the composition has more than 3 links

The method extends: one simply adds more circles and more geometric symbols ($\square, \triangle, \bigcirc, \diamondsuit, \ldots$). The final derivative is always the product of the partial derivatives of each link, with the symbols substituted at the end. The order of multiplication does not matter (the product is commutative), but the substitution must be done from the inside outwards (first $\square = x$, then $\triangle = \text{expression in } x$, and so on).

Links

Topics: Derivatives
Concepts: Composite derivative · Chain rule · Differentiation rules
Skills: Differentiating