Module 10: The chain rule and elementary derivatives

A. Prerequisites and observable goals#

B. The central idea: rates multiply along a dependency chain#

Suppose $u=g(x)$ and $y=f(u)$. A small input change first changes $u$ at rate $g'(x)$, then changes $y$ at rate $f'(u)$ per unit change in $u$. The total rate with respect to $x$ multiplies these effects. The evaluation point of the outer derivative is the inner output, not automatically $x$.

C. The rule and a compact derivative dictionary#

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then

For a differentiable inner function $u=u(x)$, useful consequences are

Here $b>0$, and the trigonometric quotient functions require their usual nonzero denominators. For arbitrary real $r$, use $u>0$ unless an appropriate real branch is separately justified. The formula for $\ln u$ itself requires $u>0$; $\ln|u|$ extends the derivative formula to nonzero negative inputs on their own intervals. These formulas are exact, not small-angle approximations. Chain-rule conditions and inverse-function derivatives: [O1Ch], [O1Inv].

D. Why the method works#

Optional proof from local linear changes. Let $u=g(x)$, let $h$ be the change in $x$, and let $k=g(x+h)-g(x)$ be the resulting change in $u$. Differentiability means that we can write the exact decompositions

where the error factors $\alpha(h)\to0$ and $\beta(k)\to0$ as their inputs approach zero. Set $\beta(0)=0$ so the second identity also covers $k=0$. For $h\ne0$, substitution gives

Differentiability of $g$ implies $k\to0$ when $h\to0$, so the limit is $f'(u)g'(x)$. This proof never divides by $k$, which could be zero. The informal “cancel $du$” mnemonic alone does not establish the theorem.

For sine, use the addition formula in its difference quotient:

The two basic radian limits make this approach $0+\cos x$. For exponentials, $e^{x+h}=e^x e^h$ reduces the derivative to $e^x\lim_{h\to0}(e^h-1)/h=e^x$. The normalization of the base $e$ makes that final limit equal to one. Other bases add the factor $\ln b$.

E. Read from outside inward, then multiply outward#

For $\sqrt{1+e^{2x}}$, the outer operation is a square root, then addition, then an exponential, then multiplication of the input by $2$. Differentiate each layer at its current input and multiply the chain factors. For $x^2e^x$, the outer operation is multiplication of two functions, so start with the product rule. A problem may need both rules: differentiate the product, then apply the chain rule within its terms.

A useful self-check is to count nonlinear nesting layers. If $\sin(5x)$ gives only $\cos(5x)$, the derivative of the inner $5x$ was lost. If a derivative of $\ln(1+x^2)$ contains $1/x$, the original inner expression was incorrectly replaced instead of preserved.

F. Worked examples with explicit layers#

Power of a trigonometric function. Write $y=[\sin(3x)]^2$. The outside square differentiates to twice its input; sine contributes cosine; $3x$ contributes $3$:

The notation $\sin^2(3x)$ means this square, not $\sin((3x)^2)$.

Product plus chain. For $y=(x^2+1)e^{-3x}$,

The exponential changes too, so treating it as a constant would lose the second term.

Finish the layer. For $y=\ln(4+x^2)$, the outer derivative is $1/(4+x^2)$. Multiply by $2x$ to finish: $y'=2x/(4+x^2)$. The log input is positive for every real $x$, so this derivative is valid everywhere.

Computing bridge: Chain rule as a dependency calculation#

For $u=3x+1$ and $y=u^2$, a small change travels through two stages: $\dd u/\dd x=3$ and $\dd y/\dd u=2u$. Their product gives $\dd y/\dd x=6(3x+1)$. A computational graph records these dependencies explicitly. T03 later applies the same rule to a loss function with many parameters. The mathematical chain rule is the foundation; a library computing gradients does not remove the need to understand the graph or the differentiability assumptions.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 10#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Return to M03.01 and M09.06, then solve M10.07 from memory of the rule rather than the numbers. Advance when you can identify the outermost operation in a fresh expression and explain the evaluation point of every chain factor. Repair composition before memorizing more derivative tables.