Differentiating term by term AB & BC

Let $f$ and $g$ be differentiable and let $c$ be a constant.

1. Constant: $\frac{d}{dx}[c] = 0$.
2. Constant multiple: $\frac{d}{dx}\left[c\,f(x)\right] = c\,f'(x)$.
3. Sum: $\frac{d}{dx}\left[f(x) + g(x)\right] = f'(x) + g'(x)$.
4. Difference: $\frac{d}{dx}\left[f(x) - g(x)\right] = f'(x) - g'(x)$.

Each follows from the corresponding limit law in Topic 1.8, applied to the difference quotient. The sum rule, for instance, is the limit of a sum being the sum of the limits, once the quotient of $f + g$ has been split into the quotient of $f$ plus the quotient of $g$.

Taken together these say that differentiation is linear. A constant multiplier is untouched and comes along for the ride; a sum is handled one piece at a time.

The practical shape of the rules is: go along the expression term by term, differentiate each one, and reassemble with the same plus and minus signs.

$$\frac{d}{dx}\left[3x^{5} - 2x^{2} + 7x - 9\right] = 15x^{4} - 4x + 7.$$

Four things happened, in order. $3x^5$ became $15x^4$, with $3 \cdot 5 = 15$. $-2x^2$ became $-4x$, keeping the minus sign. $7x$ became $7$, since $\frac{d}{dx}[x] = 1$. And $-9$ became $0$ and vanished from the answer.

The last one is where terms most often survive that should not. A constant added at the end shifts the graph vertically and changes no slope anywhere, so it contributes nothing to the derivative.

The coefficient is multiplied by the exponent and is never differentiated on its own. Treating $3x^5$ as a product of two functions and reaching for the product rule gives $0 \cdot x^5 + 3 \cdot 5x^4$, which is the same answer by a much longer road, and is where the coefficient usually gets lost.

The rules apply to a sum of terms, so an expression that is not in that form gets rewritten first. Two patterns account for most of it:

1. A single denominator splits. $\dfrac{2x^{3} + 5}{x} = 2x^{2} + 5x^{-1}$, whose derivative is $4x - 5x^{-2}$.
2. A constant denominator is a constant multiple. $\dfrac{x^{3}}{4} = \frac{1}{4}x^{3}$, whose derivative is $\frac{3}{4}x^{2}$.

Both are legal because the division being undone is by something that is either a single power of $x$ or a constant. A denominator with more than one term, such as $\dfrac{1}{x + 3}$, does not split, and waits for Topic 2.9.

Expanding a product is the same idea from the other side. $\frac{d}{dx}\left[x^2(x + 4)\right]$ can be done by multiplying out to $x^3 + 4x^2$ and differentiating term by term, giving $3x^2 + 8x$. Topic 2.8 provides a rule that avoids the expansion, but the expansion is always available and always correct.

Two things are true of sums and false of products and quotients. Getting this boundary right now prevents the two most expensive errors in the rest of the unit.

$$\frac{d}{dx}\left[f \cdot g\right] \neq f' \cdot g', \qquad \frac{d}{dx}\left[\frac{f}{g}\right] \neq \frac{f'}{g'}.$$

A one-line check kills both. Take $f(x) = x$ and $g(x) = x$. Then $f \cdot g = x^2$, whose derivative is $2x$, while $f' \cdot g' = 1 \cdot 1 = 1$. The two are not equal, so no such rule exists.

The other boundary is the one from Topic 2.5: linearity says nothing about which rule to use on an individual term. In $\frac{d}{dx}\left[x^2 + 2^x\right]$ the sum rule applies perfectly, and it hands you two separate problems, only one of which is a power.