Gilbert Strang’s Calculus: Product Rule and Quotient Rule

Calculus

Derivatives

Product Rule

Quotient Rule

Gilbert Strang

Author

Chao Ma

Published

February 23, 2026

Product

Let \[ p(x)=f(x)g(x). \] Then \[ \frac{dp}{dx}=f(x)\frac{dg}{dx}+g(x)\frac{df}{dx}. \]

Examples

$f(x)=x^2,\ g(x)=x$
Product rule gives \[ \frac{d}{dx}(x^3)=x^2+x\cdot 2x=3x^2. \]
$f(x)=x^3,\ g(x)=x$
Product rule gives \[ \frac{d}{dx}(x^4)=x^3+x\cdot 3x^2=4x^3. \]

These examples match the power-rule pattern \[ \frac{d}{dx}(x^n)=nx^{n-1}. \]

Derivative of $\frac{d}{dx}(f(x))^n$

For $n=2$: \[ \frac{d}{dx}(f(x))^2=f(x)\frac{df}{dx}+f(x)\frac{df}{dx}=2f(x)\frac{df}{dx}. \]

For $n=3$, write $f(x)^3=f(x)^2f(x)$: \[ \frac{d}{dx}(f(x)^3)=f(x)\,\frac{d}{dx}(f(x)^2)+f(x)^2\frac{df}{dx} =3f(x)^2\frac{df}{dx}. \]

So the pattern is \[ \frac{d}{dx}(f(x)^n)=n f(x)^{n-1}\frac{df}{dx}. \]

Derivative of square root

Let $f(x)=\sqrt{x}$. Since $(\sqrt{x})^2=x$: \[ \frac{d}{dx}(\sqrt{x})^2=2\sqrt{x}\,\frac{d}{dx}(\sqrt{x})=1. \] Hence \[ \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}=\frac{1}{2}x^{-1/2}, \] which also matches the power rule with $n=\frac12$.

Explanation (geometric intuition)

View $p(x)=f(x)g(x)$ as the area of a rectangle. When $x$ increases by $\Delta x$:

$\Delta f=f(x+\Delta x)-f(x)$
$\Delta g=g(x+\Delta x)-g(x)$

The area increase has three parts:

$g\,\Delta f$
$f\,\Delta g$
$\Delta f\,\Delta g$

So \[ \Delta p=f\,\Delta g+g\,\Delta f+\Delta f\,\Delta g. \] Divide by $\Delta x$: \[ \frac{\Delta p}{\Delta x}=f\frac{\Delta g}{\Delta x}+g\frac{\Delta f}{\Delta x}+\Delta g\left(\frac{\Delta f}{\Delta x}\right). \] As $\Delta x\to 0$, we have $\Delta g\to 0$, so the last term vanishes. Therefore: \[ \frac{dp}{dx}=f(x)g'(x)+g(x)f'(x). \]

Quotient

Let \[ q(x)=\frac{f(x)}{g(x)}\quad (g(x)\neq 0). \] To get $q'(x)$, start from \[ f(x)=g(x)q(x). \] Differentiate: \[ \frac{df}{dx}=g(x)\frac{dq}{dx}+q(x)\frac{dg}{dx} = g(x)\frac{dq}{dx}+\frac{f(x)}{g(x)}\frac{dg}{dx}. \] Multiply by $g(x)$: \[ g(x)\frac{df}{dx}-f(x)\frac{dg}{dx}=g(x)^2\frac{dq}{dx}. \] So \[ \frac{dq}{dx}=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}. \]

Example

Take

$f(x)=1$
$g(x)=x^n$
$q(x)=\frac{1}{x^n}=x^{-n}$

Known derivatives:

$f'(x)=0$
$g'(x)=n x^{n-1}$

Apply quotient rule: \[ q'(x)=\frac{x^n\cdot 0-1\cdot n x^{n-1}}{x^{2n}} =-n x^{-n-1}. \] This matches differentiating $x^{-n}$ directly by the power rule.

Takeaway. Product and quotient rules are not isolated formulas. They are consistent with power-rule patterns and have a clean geometric interpretation through small area changes.

--- title: "Gilbert Strang's Calculus: Product Rule and Quotient Rule" author: "Chao Ma" date: "2026-02-23" categories: [Calculus, Derivatives, Product Rule, Quotient Rule, Gilbert Strang] --- {{< video https://www.youtube.com/watch?v=5ZpqI8zz1HM&list=PLFW_V3qDH5jRyfpD9uiq6aKVTWIxpbIm3&index=7 >}} # Product Let $$ p(x)=f(x)g(x). $$ Then $$ \frac{dp}{dx}=f(x)\frac{dg}{dx}+g(x)\frac{df}{dx}. $$ ## Examples - $f(x)=x^2,\ g(x)=x$ Product rule gives $$ \frac{d}{dx}(x^3)=x^2+x\cdot 2x=3x^2. $$ - $f(x)=x^3,\ g(x)=x$ Product rule gives $$ \frac{d}{dx}(x^4)=x^3+x\cdot 3x^2=4x^3. $$ These examples match the power-rule pattern $$ \frac{d}{dx}(x^n)=nx^{n-1}. $$ ## Derivative of $\frac{d}{dx}(f(x))^n$ For $n=2$: $$ \frac{d}{dx}(f(x))^2=f(x)\frac{df}{dx}+f(x)\frac{df}{dx}=2f(x)\frac{df}{dx}. $$ For $n=3$, write $f(x)^3=f(x)^2f(x)$: $$ \frac{d}{dx}(f(x)^3)=f(x)\,\frac{d}{dx}(f(x)^2)+f(x)^2\frac{df}{dx} =3f(x)^2\frac{df}{dx}. $$ So the pattern is $$ \frac{d}{dx}(f(x)^n)=n f(x)^{n-1}\frac{df}{dx}. $$ ### Derivative of square root Let $f(x)=\sqrt{x}$. Since $(\sqrt{x})^2=x$: $$ \frac{d}{dx}(\sqrt{x})^2=2\sqrt{x}\,\frac{d}{dx}(\sqrt{x})=1. $$ Hence $$ \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}=\frac{1}{2}x^{-1/2}, $$ which also matches the power rule with $n=\frac12$. ## Explanation (geometric intuition) ![Product rule from rectangle area change](product-quotient-rule-1.png) View $p(x)=f(x)g(x)$ as the area of a rectangle. When $x$ increases by $\Delta x$: - $\Delta f=f(x+\Delta x)-f(x)$ - $\Delta g=g(x+\Delta x)-g(x)$ The area increase has three parts: - $g\,\Delta f$ - $f\,\Delta g$ - $\Delta f\,\Delta g$ So $$ \Delta p=f\,\Delta g+g\,\Delta f+\Delta f\,\Delta g. $$ Divide by $\Delta x$: $$ \frac{\Delta p}{\Delta x}=f\frac{\Delta g}{\Delta x}+g\frac{\Delta f}{\Delta x}+\Delta g\left(\frac{\Delta f}{\Delta x}\right). $$ As $\Delta x\to 0$, we have $\Delta g\to 0$, so the last term vanishes. Therefore: $$ \frac{dp}{dx}=f(x)g'(x)+g(x)f'(x). $$ # Quotient Let $$ q(x)=\frac{f(x)}{g(x)}\quad (g(x)\neq 0). $$ To get $q'(x)$, start from $$ f(x)=g(x)q(x). $$ Differentiate: $$ \frac{df}{dx}=g(x)\frac{dq}{dx}+q(x)\frac{dg}{dx} = g(x)\frac{dq}{dx}+\frac{f(x)}{g(x)}\frac{dg}{dx}. $$ Multiply by $g(x)$: $$ g(x)\frac{df}{dx}-f(x)\frac{dg}{dx}=g(x)^2\frac{dq}{dx}. $$ So $$ \frac{dq}{dx}=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}. $$ ## Example Take - $f(x)=1$ - $g(x)=x^n$ - $q(x)=\frac{1}{x^n}=x^{-n}$ Known derivatives: - $f'(x)=0$ - $g'(x)=n x^{n-1}$ Apply quotient rule: $$ q'(x)=\frac{x^n\cdot 0-1\cdot n x^{n-1}}{x^{2n}} =-n x^{-n-1}. $$ This matches differentiating $x^{-n}$ directly by the power rule. --- **Takeaway.** Product and quotient rules are not isolated formulas. They are consistent with power-rule patterns and have a clean geometric interpretation through small area changes.