Gilbert Strang’s Calculus: Chains and the Chain Rule

Calculus

Derivatives

Chain Rule

Product Rule

Exponential Function

Gilbert Strang

Author

Chao Ma

Published

February 25, 2026

Chain Function

A chain (composite) function has the form \[ f(g(x)). \]

Chain Rule

If \[ z=f(y),\quad y=g(x), \] then \[ \frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}. \]

Examples

For \[ f(x)=\sin(3x): \]

$z=\sin y$
$y=3x$

\[ \frac{dz}{dy}=\cos y,\qquad \frac{dy}{dx}=3, \] so \[ \frac{dz}{dx}=3\cos y=3\cos(3x). \]

For \[ f(x)=(x^3)^2: \]

$z=y^2$
$y=x^3$

\[ \frac{dz}{dy}=2y,\qquad \frac{dy}{dx}=3x^2, \] therefore \[ \frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}=2y\cdot 3x^2=6x^5. \]

Another example: \[ f(x)=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}. \]

$z=y^{-1/2}$
$y=1-x^2$

\[ \frac{dz}{dy}=-\frac{1}{2}y^{-3/2},\qquad \frac{dy}{dx}=-2x, \] so \[ \frac{dz}{dx}=\left(-\frac{1}{2}y^{-3/2}\right)(-2x)=x(1-x^2)^{-3/2}. \]

Why the Chain Rule works

Write the overall rate with an intermediate step: \[ \frac{\Delta z}{\Delta x}=\frac{\Delta z}{\Delta y}\cdot\frac{\Delta y}{\Delta x}. \]

This is exactly the “break-down” idea: a nested function changes in two stages, and total sensitivity is the product of stage sensitivities.

As $\Delta x\to 0$, these difference quotients become derivatives: \[ \frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}. \]

Chained Exponential Function

Take \[ f(x)=e^{-x^2/2}. \] Set - $y=-\frac{x^2}{2}$ - $z=e^y$

Then \[ \frac{dz}{dy}=e^y,\qquad \frac{dy}{dx}=-x, \] so \[ \frac{dz}{dx}=e^y(-x)=-x e^{-x^2/2}. \]

Bell-shaped function and first derivative intuition

As the bell-shaped curve rises to its peak and then decays, the slope moves from positive to zero (at the top) and then negative.

Why are the maximum and minimum of $z'(x)$ at $x=1$ and $x=-1$? Because $z''(x)=0$ at these two points.

Now compute the second derivative: \[ \frac{dz}{dx}=-x e^{-x^2/2}. \] Use product rule with - $a(x)=-x$ - $b(x)=e^{-x^2/2}$

Then \[ a'(x)=-1,\qquad b'(x)=-x e^{-x^2/2}, \] and \[ \frac{d^2z}{dx^2}=a(x)b'(x)+a'(x)b(x) = (-x)(-x e^{-x^2/2})+(-1)e^{-x^2/2} = (x^2-1)e^{-x^2/2}. \]

Second derivative shape and concavity change

The second derivative explains where curvature changes: central concave-down behavior with concave-up behavior in outer regions.

Takeaway. Chain rule decomposes complicated derivatives into a sequence of simpler derivatives. Combined with product rule, it gives clean first- and second-derivative formulas for important functions like the Gaussian shape $e^{-x^2/2}$.

--- title: "Gilbert Strang's Calculus: Chains and the Chain Rule" author: "Chao Ma" date: "2026-02-25" categories: [Calculus, Derivatives, Chain Rule, Product Rule, Exponential Function, Gilbert Strang] --- {{< video https://www.youtube.com/watch?v=yQrKXo89nHA&list=PLFW_V3qDH5jRyfpD9uiq6aKVTWIxpbIm3&index=8 >}} # Chain Function A chain (composite) function has the form $$ f(g(x)). $$ # Chain Rule If $$ z=f(y),\quad y=g(x), $$ then $$ \frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}. $$ ## Examples For $$ f(x)=\sin(3x): $$ - $z=\sin y$ - $y=3x$ $$ \frac{dz}{dy}=\cos y,\qquad \frac{dy}{dx}=3, $$ so $$ \frac{dz}{dx}=3\cos y=3\cos(3x). $$ For $$ f(x)=(x^3)^2: $$ - $z=y^2$ - $y=x^3$ $$ \frac{dz}{dy}=2y,\qquad \frac{dy}{dx}=3x^2, $$ therefore $$ \frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}=2y\cdot 3x^2=6x^5. $$ Another example: $$ f(x)=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}. $$ - $z=y^{-1/2}$ - $y=1-x^2$ $$ \frac{dz}{dy}=-\frac{1}{2}y^{-3/2},\qquad \frac{dy}{dx}=-2x, $$ so $$ \frac{dz}{dx}=\left(-\frac{1}{2}y^{-3/2}\right)(-2x)=x(1-x^2)^{-3/2}. $$ # Why the Chain Rule works Write the overall rate with an intermediate step: $$ \frac{\Delta z}{\Delta x}=\frac{\Delta z}{\Delta y}\cdot\frac{\Delta y}{\Delta x}. $$ This is exactly the "break-down" idea: a nested function changes in two stages, and total sensitivity is the product of stage sensitivities. As $\Delta x\to 0$, these difference quotients become derivatives: $$ \frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}. $$ # Chained Exponential Function Take $$ f(x)=e^{-x^2/2}. $$ Set - $y=-\frac{x^2}{2}$ - $z=e^y$ Then $$ \frac{dz}{dy}=e^y,\qquad \frac{dy}{dx}=-x, $$ so $$ \frac{dz}{dx}=e^y(-x)=-x e^{-x^2/2}. $$ ![Bell-shaped function and first derivative intuition](chain-rule-1.png) As the bell-shaped curve rises to its peak and then decays, the slope moves from positive to zero (at the top) and then negative. Why are the maximum and minimum of $z'(x)$ at $x=1$ and $x=-1$? Because $z''(x)=0$ at these two points. Now compute the second derivative: $$ \frac{dz}{dx}=-x e^{-x^2/2}. $$ Use product rule with - $a(x)=-x$ - $b(x)=e^{-x^2/2}$ Then $$ a'(x)=-1,\qquad b'(x)=-x e^{-x^2/2}, $$ and $$ \frac{d^2z}{dx^2}=a(x)b'(x)+a'(x)b(x) = (-x)(-x e^{-x^2/2})+(-1)e^{-x^2/2} = (x^2-1)e^{-x^2/2}. $$ ![Second derivative shape and concavity change](chain-rule-2.png) The second derivative explains where curvature changes: central concave-down behavior with concave-up behavior in outer regions. --- **Takeaway.** Chain rule decomposes complicated derivatives into a sequence of simpler derivatives. Combined with product rule, it gives clean first- and second-derivative formulas for important functions like the Gaussian shape $e^{-x^2/2}$.