MISSION · Calculus

The Rules for Differentiating

The shortcuts that replace the limit, and when each applies

Explorer level
Grades 11-12
Quest time
8 minutes
Mission skill
Differentiating without going back to first principles

LEVEL 1 · EXPLORE

Make the idea move

Change the controls and watch what happens. Follow your curiosity and see what you notice.

Loading the activity…

LEVEL 2 · LEARN

Unlock the explanation

Open the chapters in any order. Try the examples as you go.

The idea in one sentence

Every derivative could be found from the limit definition, and almost none are — because a handful of rules cover everything you’ll meet.

The power rule

\[\frac{d}{dx}x^n = n x^{n-1}\]

Bring the power down in front, then reduce it by one. \(x^5\) becomes \(5x^4\); \(x^2\) becomes \(2x\); \(x\) becomes 1.

It works for every exponent, not just whole ones. \(\sqrt{x} = x^{1/2}\) differentiates to \(\frac{1}{2}x^{-1/2}\), and \(\frac{1}{x} = x^{-1}\) becomes \(-x^{-2}\). Rewriting roots and fractions as powers first is the standard opening move.

A constant differentiates to zero, which the rule delivers if you write \(7\) as \(7x^0\). Geometrically it’s obvious: a constant function is a flat line, and flat means slope zero.

Sums, constants, products, quotients

Sums come apart term by term: differentiate each piece and add. So \(3x^4 - 2x + 7\) becomes \(12x^3 - 2\).

Constant multiples come along for the ride: \(\frac{d}{dx}[5f(x)] = 5f'(x)\).

Products do not. \(\frac{d}{dx}[f \cdot g]\) is not \(f' \cdot g'\) — a tempting symmetry that’s simply false. The correct rule:

\[(fg)' = f'g + fg'\]

Quotients follow from it:

\[\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\]

The order matters in the numerator — subtraction isn’t symmetric, and swapping the terms flips the sign of every answer.

The chain rule

For a function inside a function — \((3x + 1)^5\), \(\sin(x^2)\), \(e^{4x}\) — differentiate the outside, keeping the inside untouched, then multiply by the derivative of the inside.

\[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\]

So \((3x+1)^5\) gives \(5(3x+1)^4 \cdot 3 = 15(3x+1)^4\). That final \(\times 3\) is the step everyone forgets, and it’s the whole content of the rule.

The name describes the mechanism: rates multiply along a chain. If \(y\) changes 5 times as fast as \(u\), and \(u\) changes 3 times as fast as \(x\), then \(y\) changes 15 times as fast as \(x\).

Choosing the right rule

Look at the structure before differentiating anything:

  • Terms added or subtracted → handle each separately.
  • Two things multiplied → product rule.
  • One thing over another → quotient rule.
  • Something wrapped inside something else → chain rule.

Complicated expressions need several, applied from the outside in. The most common failure isn’t arithmetic, it’s misreading which structure you’re looking at.

Common mix-ups

  • Differentiating a product as \(f'g'\). It isn’t.
  • Dropping the inside derivative in the chain rule.
  • Flipping the quotient rule’s numerator, which negates the answer.
  • Forgetting to rewrite roots as powers before applying the power rule.

Check yourself

Differentiate \(y = (2x^2 + 1)^3\) and \(y = x^2 \sin x\).

Chain rule: \(3(2x^2+1)^2 \cdot 4x = 12x(2x^2+1)^2\). Product rule: \(2x\sin x + x^2\cos x\).

LEVEL 3 · REFLECT

Add this quest to your trail

When you feel ready, mark this idea as explored. You can always come back and try it again.

EXPLORER STRATEGIES

More ways to see it

Loading shared methods…
Share how you learned it