The basic differentiation rules are short. The harder skill is deciding which rule fits the expression in front of you. A product does not always need the product rule, a quotient does not always need the quotient rule, and some expressions are much easier after a small algebraic rewrite.

It is worth separating two questions. First, what rule is valid? Second, what approach is simplest?

The core rules

For a power, \(\tfrac{d}{dx}x^r=rx^{r-1}\), where the expression is defined. Sums, differences, and constant multiples can be differentiated term by term. The basic trigonometric, exponential, and logarithmic derivatives include

\[\frac{d}{dx}\sin x=\cos x \qquad \frac{d}{dx}\cos x=-\sin x \qquad \frac{d}{dx}e^x=e^x \qquad \frac{d}{dx}\ln x=\frac1x\]

For products, \((fg)'(x)=f'(x)g(x)+f(x)g'(x)\), and for quotients, when \(g(x)\neq0\),

\[\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\]

These rules assume the relevant component functions are differentiable at the point, and if that condition fails, the rule cannot simply be applied mechanically.

Rewrite before using a longer rule

The product and quotient rules are often correct even when they are unnecessary. The drill below asks only for the first move.

Ten expressions, cycling. Five of them want a rewrite, and those are the ones worth arguing about: two products that collapse to a single power, one quotient that is really a division, and the two trigonometric functions the framework says to rearrange rather than memorize. The product and quotient rules would give the right answer on all five; they are just the long way round, and the long way is where sign errors live.

Consider \(\tfrac{x^3+2x}{x}\). The quotient rule works, but simplifying first is much easier, since \(\tfrac{x^3+2x}{x}=x^2+2\) for \(x\neq0\), and then \(\tfrac{d}{dx}(x^2+2)=2x\). Using the quotient rule produces the same derivative with more algebra and more opportunities for error.

The same idea applies to products that collapse to a single power or rational expressions that simplify before differentiation. Before choosing a rule, ask whether ordinary algebra makes the structure simpler.

Deriving the other trigonometric rules

Sine and cosine are enough to derive the other standard trigonometric derivatives. For example, \(\tan x=\tfrac{\sin x}{\cos x}\), so using the quotient rule,

\[(\tan x)' = \frac{\cos^2x+\sin^2x}{\cos^2x} = \frac{1}{\cos^2x} = \sec^2x\]

Likewise, \(\sec x=\tfrac1{\cos x}\) can be differentiated using the quotient rule or a negative power. This is useful even if you eventually memorize the formulas, because knowing where they come from makes them easier to recover when memory fails.

Recognizing a derivative inside a limit

The same classification skill can run in reverse. Consider

\[\lim_{h\to0} \frac{(2+h)^5-2^5}{h}\]

Direct substitution gives \(0/0\). You could expand \((2+h)^5\), cancel \(h\), and then evaluate the limit, but the expression is already the definition of a derivative.

If \(f(x)=x^5\), then the limit is \(f'(2)\), and since \(f'(x)=5x^4\), the answer is \(f'(2)=5(2)^4=80\). Two features reveal the structure:

  • the limit is taken as \(h\to0\)
  • the numerator has the form \(f(a+h)-f(a)\)

Recognizing the form is the entire problem.

A useful practice exercise is to take a page of derivative problems and write only the first move beside each one. Do not differentiate yet. For each expression, decide whether you would simplify, use a basic derivative, use the product rule, use the quotient rule, or use the chain rule. That isolates the classification skill from the algebra that follows.