Differentiation Rules

Computing every derivative directly from Definition 10.2's limit — "from first principle" — is correct but, as the book notes, "extremely laborious and difficult... in the majority of cases." The practical fix is to prove, once and for all, a small set of general rules that let any combination of known derivatives be differentiated automatically, with no limit computation needed at the point of use. This section proves the five rules that make differentiation an algebraic, rule-following process rather than a fresh limit every time: differentiating a sum, a product, a quotient, a composite (chain rule), and a constant multiple.

Throughout, $u=u(x)$ and $v=v(x)$ denote differentiable functions of $x$ on a common interval $I$.

Theorem 10.2 (Sum Rule). $\dfrac{d}{dx}(u+v) = \dfrac{du}{dx}+\dfrac{dv}{dx}$.

Proof idea. With $y=f(x)=u(x)+v(x)$, $f(x+\Delta x)-f(x) = \big[u(x+\Delta x)-u(x)\big]+\big[v(x+\Delta x)-v(x)\big]$, so dividing by $\Delta x$ and taking the limit term-by-term (each limit exists by hypothesis) gives $f'(x)=u'(x)+v'(x)$. The rule extends by repeated application to any finite sum: $(u_1+u_2+\cdots+u_n)' = u_1'+u_2'+\cdots+u_n'$.

Theorem 10.3 (Product Rule). $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx}+v\dfrac{du}{dx}$, i.e. $(uv)'=uv'+vu'$.

Proof idea. Write $f(x)=u(x)v(x)$. The key algebraic trick is to insert and subtract $u(x)v(x+\Delta x)$ inside $f(x+\Delta x)-f(x)$:

Dividing by $\Delta x$ and letting $\Delta x\to0$: the first term's difference quotient $\to u'(x)$ while $v(x+\Delta x)\to v(x)$ (since $v$, being differentiable, is continuous by Theorem 10.1), and the second term's difference quotient $\to v'(x)$ with $u(x)$ fixed — giving $f'(x) = v(x)u'(x)+u(x)v'(x)$. By repeated use, $(uvw)'=u'vw+uv'w+uvw'$, and in general the derivative of a product of $n$ differentiable functions is the sum of $n$ terms, each formed by differentiating exactly one factor and leaving the rest untouched.

Theorem 10.4 (Quotient Rule). If $v(x)\ne0$, $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\,\dfrac{du}{dx}-u\,\dfrac{dv}{dx}}{v^2}$, i.e. $\left(\dfrac{u}{v}\right)' = \dfrac{vu'-uv'}{v^2}$.

Proof idea. With $f=u/v$, the same insert-and-subtract trick on $f(x+\Delta x)-f(x)=\dfrac{u(x+\Delta x)}{v(x+\Delta x)}-\dfrac{u(x)}{v(x)}$ produces, after combining over a common denominator $v(x)v(x+\Delta x)$,

Letting $\Delta x\to0$ (again using continuity of $v$ so $v(x+\Delta x)\to v(x)$) gives $f'(x) = \dfrac{v(x)u'(x)-u(x)v'(x)}{[v(x)]^2}$.

Theorem 10.5 (Chain Rule). If $y=f(u)$ is a differentiable function of $u$, and $u=g(x)$ is a differentiable function of $x$, so that $y=f(g(x))=(f\circ g)(x)$, then

$\frac{d}{dx}\big[f(g(x))\big] = f'(g(x))\,g'(x).$

Here $u=g(x)$ is called the inner function and $f$ the outer function; $u$ itself is called the intermediate argument. Proof idea. Writing $\Delta u = g(x+\Delta x)-g(x)$ (so $\Delta u\to0$ as $\Delta x\to0$, since $g$ is continuous), the difference quotient splits as a product,

$\frac{\Delta y}{\Delta x} = \frac{f(u+\Delta u)-f(u)}{\Delta u}\times\frac{g(x+\Delta x)-g(x)}{\Delta x} = \frac{\Delta y}{\Delta u}\times\frac{\Delta u}{\Delta x},$ …