Fundamental Differentiation Rules

Quick Differentiation Rules

When learning differentiation, it helps to think in terms of how change propagates rather than just memorising formulas.

Product Rule

When two functions are multiplied, both are changing at the same time, so the total change comes from two sources:

$ \frac{d}{dx}(uv) = u’v + uv’ $

Intuition

• One term captures $u$ changing while $v$ stays fixed
• The other captures $v$ changing while $u$ stays fixed

Quotient Rule

The quotient rule can be seen as a natural extension of the product rule combined with the chain rule:

$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u’v - uv’}{v^2} $

Intuition

• Rewrite the quotient as $u \cdot v^{-1}$
• Then apply the product rule and chain rule

Chain Rule

The chain rule describes how change flows through multiple layers:

$ \frac{d}{dx} f(g(x)) = f’(g(x)) \cdot g’(x) $

Intuition

• Change happens in two stages
  1. The inner function $g(x)$ changes
  2. The outer function $f(\cdot)$ responds to that change

Why These Rules Work

1) Why does the Product Rule work?

$ (uv)’ = u’v + uv’ $

Core idea: both functions change, so there are two sources of change.

Suppose we have two functions $u(x)$ and $v(x)$.
If we move by a small amount $dx$, then:

• $u$ changes
• $v$ also changes

So the total change in the product is:

$ \Delta(uv) = (u + \Delta u)(v + \Delta v) - uv $

Expanding this gives:

$ uv + u\Delta v + v\Delta u + \Delta u\Delta v - uv $

The $uv$ terms cancel, leaving:

$ u\Delta v + v\Delta u + \Delta u\Delta v $

Now, $\Delta u\Delta v$ is a tiny value multiplied by another tiny value, so it becomes negligible.

So we keep:

$ u\Delta v + v\Delta u $

Dividing both sides by $dx$:

$ \frac{\Delta(uv)}{dx} = u\frac{\Delta v}{dx} + v\frac{\Delta u}{dx} $

Taking the limit gives:

$ (uv)’ = u’v + uv’ $

2) Why does the Quotient Rule work?

$ \left(\frac{u}{v}\right)’ = \frac{u’v - uv’}{v^2} $

Core idea: a quotient can be rewritten as a product with a reciprocal.

$ \frac{u}{v} = u \cdot v^{-1} $

Differentiate using the product rule:

$ \frac{d}{dx}(u v^{-1}) = u(v^{-1})’ + u’v^{-1} $

Now differentiate $v^{-1}$ using the chain rule:

$ (v^{-1})’ = -v^{-2}v’ $

Substitute back:

$ u(-v^{-2}v’) + u’v^{-1} $

Rearranging:

$ \frac{u’v - uv’}{v^2} $

So the quotient rule is really just the product rule + chain rule.

3) Why does the Chain Rule work?

$ \frac{d}{dx}f(g(x)) = f’(g(x)) \cdot g’(x) $

Core idea: change happens in two stages.

• If $x$ changes, then $g(x)$ changes
• If $g(x)$ changes, then $f(g(x))$ changes

So the rate of change is passed along like a chain.

Using small changes:

$ \Delta f = f’(g) \cdot \Delta g $

and

$ \Delta g = g’(x) \cdot \Delta x $

Combine them:

$ \Delta f = f’(g) \cdot g’(x) \cdot \Delta x $

Now divide by $\Delta x$:

$ \frac{\Delta f}{\Delta x} = f’(g(x))g’(x) $

Taking the limit gives:

$ \frac{df}{dx} = f’(g(x))g’(x) $

Summary

• Product rule → both parts change, so two change terms appear
• Quotient rule → comes from rewriting as a product with a reciprocal
• Chain rule → change propagates through layers

Once you see the intuition, these rules feel much more natural than just memorising formulas.