q(t) = (t² + 2t + 1) sin(t²) requires product: (2t + 2) sin(t²) + (t² + 2t + 1) (2t cos(t²)). Chain for sin(t²). Error: Missing chain, like A. Strategy: Composite sin needs chain in product.

Since h(t) is the product of sin(t) and (t² + 1), the product rule is required: derivative is cos(t)(t² + 1) + sin(t)(2t). This captures the rate of change from both factors. Without it, one might wrongly differentiate only the trigonometric part, yielding just cos(t)(t² + 1) as in C. Another shortcut error is adding derivatives without multiplying, like choice A. The full application avoids these pitfalls. For transferable strategy, scan for multiplication of two non-constant functions and apply the product rule to ensure complete differentiation.

D(p) = (p + 1) $e^{p²}$ needs product: 1 · $e^{p²}$ + (p + 1) (2p $e^{p²}$). Chain for exponent. Shortcut: Missing chain, like C. Strategy: Exponential with variable power uses chain in product.

For b(x) = (x² + 1) sin²(x), the product rule yields 2x sin²(x) + (x² + 1)(2 sin(x) cos(x)). It's required as a product of a polynomial and a squared sine. The derivative of sin²(x) is 2 sin(x) cos(x) via chain rule. Tempting shortcuts include forgetting the chain or one term, as in choice D. Always include both. Check for product form; if present, use the rule and chain where needed for transferable accuracy.

c(x) = $(x + 2) \sqrt{x + 1}$ requires product rule: $1 \cdot \sqrt{x + 1} + (x + 2) \left( \frac{1}{2 \sqrt{x + 1}} \right)$. Combines linear and root derivatives. Shortcut: Ignoring root derivative, like D. Others add without product. Strategy: For root products, apply rule with chain for roots.

We need the product rule here because $f(x)$ is the product of $(x^3-1)$ and $e^x$. The product rule formula is: $(uv)' = u'v + uv'$. Let $u = x^3-1$ with $u' = 3x^2$, and $v = e^x$ with $v' = e^x$. Applying the product rule: $f'(x) = 3x^2 \cdot e^x + (x^3-1) \cdot e^x$. Notice that we can factor out $e^x$ to get $f'(x) = (3x^2 + x^3 - 1)e^x$, but the unfactored form $3x^2e^x + (x^3-1)e^x$ is also correct. A tempting error is to just multiply the derivatives $3x^2 \cdot e^x$, forgetting the second term. When differentiating products, always remember both terms: derivative of first times second, plus first times derivative of second.

The function $q(x) = (\tan x)(x^2)$ requires the product rule because it's a product of two functions. We have $u = \tan x$ with $u' = \sec^2 x$, and $v = x^2$ with $v' = 2x$. Applying the product rule: $q'(x) = \sec^2 x \cdot x^2 + \tan x \cdot 2x$. This gives us $q'(x) = x^2\sec^2 x + 2x\tan x$. A tempting shortcut might be to just multiply the derivatives: $\sec^2 x \cdot 2x$, but this misses the essential structure of the product rule. When you see any product of functions—even when one seems "simple" like $x^2$—you must apply the full product rule formula to capture both required terms.

g(t) = √t cos(2t) is product: (1/(2√t)) cos(2t) + √t (-2 sin(2t)), or (1/(2√t)) cos(2t) - 2 √t sin(2t). Chain for cos(2t). But wait, verifying: derivative of cos(2t) is -sin(2t) * 2 = -2 sin(2t), yes. Marked A is correct. Tempting to miss the 2 in chain, like B. Strategy: Double chain in products.

Q(x) = (x² - 4x + 4) $e^{-x}$ needs product: (2x - $4)e^{-x}$ - (x² - 4x + $4)e^{-x}$. For quadratic-exponential. Exponential: $-e^{-x}$. Tempting: only first, choice B. Full rule. Apply to polys with exponentials.

V(x) = $(x^2 + 3x + 1) \arccos(x)$ uses product: $(2x + 3) \arccos(x) - \frac{(x^2 + 3x + 1)}{\sqrt{1 - x^2}}$. Quadratic-arccos. Arccos: $-\frac{1}{\sqrt{1 - x^2}}$. Shortcut: ignore sign, but correct. Full rule. Apply to polys with inverse trig.