Derivative Rules

Why we need “rules”

You already know what a derivative means: the instantaneous rate of change, or the slope of the tangent line. The definition is

f′(x) = lim_{h→0} (f(x+h) − f(x)) / h.

That definition is the foundation—but using it from scratch for every function is slow. In practice, functions are built from simpler ones, and we want a fast way to translate “how the function is built” into “how its slope behaves.”

Derivative rules are exactly that translation layer.

What counts as a “rule”?

A derivative rule is a statement of the form

If a function has structure X, then its derivative has structure Y.

For example:

• •If the function is a power xⁿ, its derivative is n·xⁿ⁻¹.
• •If the function is a product f(x)g(x), its derivative is f′g + fg′.
• •If the function is a composition f(g(x)), its derivative is f′(g(x))·g′(x).

Operator viewpoint (useful habit)

Write the derivative as an operator:

D = d/dx.

Then rules can be read as “how D interacts with algebra.” For instance:

• •D[xⁿ] = n·xⁿ⁻¹
• •D[f·g] = (Df)·g + f·(Dg)

This operator view helps you avoid a common mistake: differentiating “piece by piece” incorrectly when multiplication or composition is involved.

Domain and “where defined”

Some rules mention conditions like “where defined.” Example: xⁿ with real n isn’t always defined for negative x (e.g., x^{1/2}). When we say

d/dx (xⁿ) = n·xⁿ⁻¹,

we mean: on any interval where xⁿ is defined as a real-valued differentiable function, this derivative formula holds.

Big picture

In early calculus, the workflow is:

1. 1)Parse the structure (power, product, quotient, composition).
2. 2)Apply the matching derivative rule.
3. 3)Simplify.

The rest of this lesson builds that skill carefully, one structure at a time.

Why the power rule matters

Polynomials are the “atoms” of algebraic functions. If you can differentiate xⁿ quickly, you can differentiate any polynomial, and then many more functions after combining with other rules.

Power rule statement

For real n (where xⁿ is differentiable),

d/dx (xⁿ) = n·xⁿ⁻¹.

Examples:

• •d/dx (x⁵) = 5x⁴
• •d/dx (x) = d/dx (x¹) = 1·x⁰ = 1
• •d/dx (1) = d/dx (x⁰) = 0·x⁻¹ = 0 (consistent: derivative of a constant is 0)

Constant multiple rule (used constantly)

If c is a constant and f is differentiable, then

d/dx (c·f(x)) = c·f′(x).

This means you can “pull constants out” before differentiating.

Linear combinations (bonus that you’ll use implicitly)

If f and g are differentiable, then

d/dx (f(x) + g(x)) = f′(x) + g′(x),

d/dx (f(x) − g(x)) = f′(x) − g′(x).

These combine with the power rule to make polynomials immediate.

Power rule for negative and fractional powers

The formula still works, but you must respect domains.

1) Negative powers (x ≠ 0):

f(x) = x^{−3} = 1/x³

f′(x) = −3·x^{−4} = −3/x⁴.

2) Fractional powers (domain depends):

f(x) = x^{1/2} = √x (domain x ≥ 0)

f′(x) = (1/2)·x^{−1/2} = 1/(2√x),

valid for x > 0 (note: derivative blows up at x = 0).

A short “show your work” derivation (intuition)

A key reason the power rule is believable is how it behaves for integer n. For n = 2:

Let f(x) = x².

f′(x) = lim_{h→0} ((x+h)² − x²)/h

= lim_{h→0} (x² + 2xh + h² − x²)/h

= lim_{h→0} (2xh + h²)/h

= lim_{h→0} (2x + h)

= 2x.

That matches 2·x¹.

When the power rule is not enough

If you see something like (3x² + 1)⁵, the outside is a power, but the inside is not just x. That is a composition, and you’ll need the chain rule later in this lesson.

For now, power rule handles expressions where the variable is directly the base (xⁿ), plus sums and constant multiples.

Why products and quotients need special care

A tempting (but wrong) idea is:

(d/dx)(f·g) ?= f′·g′.

That fails because when x changes a little, both factors change, and their interaction matters. The product rule accounts for that interaction.

Product rule

Statement

If f and g are differentiable, then

d/dx (f(x)g(x)) = f′(x)g(x) + f(x)g′(x).

Why it looks like that (tiny-change intuition)

Let x increase by a small amount Δx. Then:

• •f changes by about f′(x)Δx
• •g changes by about g′(x)Δx

The product changes like:

(f + Δf)(g + Δg) − fg

= fΔg + gΔf + ΔfΔg.

Divide by Δx:

(fΔg)/Δx + (gΔf)/Δx + (ΔfΔg)/Δx.

As Δx → 0:

• •Δg/Δx → g′
• •Δf/Δx → f′
• •(ΔfΔg)/Δx → 0 (it’s “second order small”)

So the derivative becomes f·g′ + g·f′.

Practical pattern

For (something)·(something), write:

(fg)′ = f′g + fg′.

Differentiate the first factor, keep the second the same, then swap.

Quotient rule

Statement

If f and g are differentiable and g(x) ≠ 0, then

d/dx (f(x)/g(x)) = (f′(x)g(x) − f(x)g′(x)) / (g(x))².

Why it’s true (derive from product + chain)

Instead of memorizing, you can rebuild it:

f/g = f · (g^{−1}).

Differentiate using product rule:

d/dx [f·g^{−1}] = f′·g^{−1} + f·d/dx[g^{−1}].

Now use chain rule on g^{−1} (we’ll formalize chain rule soon, but we can still track structure):

d/dx[g^{−1}] = (−1)·g^{−2}·g′.

So:

(f/g)′ = f′·g^{−1} + f·(−1)·g^{−2}·g′

= f′/g − f g′/g²

= (f′g)/g² − (f g′)/g²

= (f′g − f g′)/g².

Product vs quotient: when to use which?

If you see a division bar, quotient rule works. But often it’s simpler to rewrite as a product with a power:

(3x² + 1)/(x⁴) = (3x² + 1)·x^{−4}.

Then use product rule + power rule.

Here’s a quick comparison:

Breathing room: a reality check

• •Product rule adds two terms.
• •Quotient rule creates a numerator difference and a squared denominator.

This “expression growth” is normal. After differentiating, simplify carefully (factor common pieces, combine like terms). Don’t try to simplify during differentiation unless it’s obviously helpful.

Why the chain rule is the most important rule

Many functions are nested: you don’t just compute something from x, you compute something from something-from-x.

Examples:

• •(3x + 1)⁵
• •√(1 + x²)
• •(sin(x))² (if you already know trig derivatives later)

In all of these, there is an outside function applied to an inside function.

Composition language

If y = f(u) and u = g(x), then y = f(g(x)).

The chain rule connects their rates of change:

If x changes, u changes; if u changes, y changes.

Statement

If y = f(g(x)) and both functions are differentiable (where needed), then

d/dx f(g(x)) = f′(g(x)) · g′(x).

Read it as:

• •Differentiate the outside function (as if its input were “a blank”).
• •Then multiply by the derivative of the inside.

Why it works (rate-of-change viewpoint)

Think of derivatives as “rates”:

dy/dx = (dy/du)·(du/dx).

This is not just a mnemonic; it reflects that small changes propagate:

• •x changes by dx
• •that causes u to change by du ≈ (du/dx)dx
• •that causes y to change by dy ≈ (dy/du)du

Combine:

dy ≈ (dy/du)·(du/dx)·dx,

so dy/dx ≈ (dy/du)(du/dx), and in the limit it becomes exact.

The most common chain-rule forms

1) Power of a function:

d/dx (g(x))ⁿ = n·(g(x))ⁿ⁻¹ · g′(x).

2) Root of a function:

d/dx √(g(x)) = d/dx (g(x))^{1/2}

= (1/2)(g(x))^{−1/2}·g′(x)

= g′(x)/(2√(g(x))).

Identifying “inside” and “outside”

Practice saying it out loud:

• •For (3x + 1)⁵: outside is u⁵, inside is u = 3x + 1.
• •For √(1 + x²): outside is √u, inside is u = 1 + x².

Multi-layer nesting

Sometimes you have multiple layers:

h(x) = (1 + (2x − 3)²)^{5}.

You can name layers:

• •a(x) = 2x − 3
• •b(x) = 1 + (a(x))²
• •h(x) = (b(x))⁵

Then apply chain rule repeatedly:

h′(x) = 5(b(x))⁴ · b′(x)

b′(x) = 2a(x) · a′(x)

a′(x) = 2.

Finally substitute back.

Product/quotient plus chain is the “real world” mix

Most nontrivial derivatives are combinations. A good strategy is:

1. 1)Find the outermost operation (sum, product, quotient, composition).
2. 2)Apply its rule.
3. 3)Recursively differentiate the pieces.

Take a breath when expressions get big: the rules are local and mechanical, even when the algebra looks messy.

Putting the rules together (workflow)

Most functions you’ll differentiate now are built from:

• •sums and constants,
• •powers,
• •products,
• •quotients,
• •compositions.

A reliable workflow:

1) Rewrite to clarify structure.

• •Replace √(…) with (…)^{1/2}
• •Replace 1/(…) with (…)^{−1}

2) Mark the top-level structure.

• •Is it A + B? A·B? A/B? f(g(x))?

3) Differentiate using the matching rule.

• •Apply product/quotient/chain at the top.

4) Simplify.

• •Factor common terms.
• •Combine fractions.
• •Reduce powers.

Why simplification is not just “cosmetic”

Simplifying can reveal:

• •where the derivative is undefined,
• •zeros of the derivative (critical points),
• •cancellations that prevent numerical blow-ups.

How this connects to later topics

These rules are prerequisites because they scale.

1) Multivariable calculus

When functions depend on multiple variables, you’ll use partial derivatives like ∂/∂x. The algebraic structure rules (product rule, chain rule) still apply, but with partials.

Connection: Multivariable Calculus

2) Multivariable chain rule

In multiple dimensions, compositions become richer (e.g., y = f(g(x)) where g outputs a vector). The idea “outside derivative times inside derivative” becomes Jacobians and matrix multiplication.

Connection: Multivariable Chain Rule

3) Taylor series

Taylor series uses higher derivatives f″(x), f‴(x), etc. Computing those efficiently depends on being fluent with these first-derivative rules; otherwise higher derivatives become impossible to manage.

Connection: Taylor Series

A quick comparison table (what each rule reacts to)

Final mindset

Derivative rules are not a bag of tricks—they’re a language. Once you can “read” an expression’s structure, differentiation becomes systematic and predictable.