This problem requires the product rule to differentiate g(x) = arctan(x)(x³ - 2). The product rule states that for f(x) = u(x)·v(x), we have f'(x) = u'(x)·v(x) + u(x)·v'(x). Here, u(x) = arctan(x) with u'(x) = 1/(1 + x²), and v(x) = x³ - 2 with v'(x) = 3x². Applying the product rule: g'(x) = (1/(1 + x²))·(x³ - 2) + arctan(x)·3x² = (x³ - 2)/(1 + x²) + 3x²arctan(x). Choice A shows only the first term, missing the crucial second product of arctan(x) and 3x². The product rule always yields two terms—both must be included to obtain the correct derivative.

This problem requires the product rule to find the derivative of a volume function given as a product of two functions. To apply the product rule, identify u(r) = 4r³ with u'(r) = 12r², and v(r) = arctan r with v'(r) = 1/(1 + r²). Then, the derivative V'(r) is u'(r)v(r) + u(r)v'(r), which expands to 12r² arctan r + 4r³ [1/(1 + r²)]. This matches choice C, providing the complete derivative. A tempting distractor like choice A omits the second term, failing to account for the derivative of the arctan function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.

This problem requires the product rule to find the derivative of a temperature model given as a product of two functions. To apply the product rule, identify u(x) = $x^2 - 2x$ with u'(x) = $2x - 2$, and v(x) = $e^{3x}$ with v'(x) = $3e^{3x}$. Then, the derivative T'(x) is $u'(x)v(x) + u(x)v'(x)$, which expands to $(2x - 2)e^{3x} + (x^2 - 2x)(3e^{3x})$. This matches choice A, providing the complete derivative. A tempting distractor like choice B omits the second term, failing to include the derivative of the exponential function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.

This problem requires the product rule to differentiate A(x) = sin x · (x³ - 1). The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = sin x with u'(x) = cos x, and v(x) = x³ - 1 with v'(x) = 3x². Applying the product rule: A'(x) = (cos x)(x³ - 1) + (sin x)(3x²) = cos x(x³ - 1) + sin x(3x²). A common mistake is to compute only cos x(3x²) (choice B), which incorrectly pairs the derivative of sin x with the derivative of (x³ - 1). The product rule pairs each function with the derivative of the other—not derivative with derivative.

This problem requires the product rule to differentiate T(t) = eᵗcos t. The product rule states that if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = eᵗ with u'(t) = eᵗ, and v(t) = cos t with v'(t) = -sin t. Applying the product rule: T'(t) = (eᵗ)(cos t) + (eᵗ)(-sin t) = eᵗcos t - eᵗsin t. A common mistake is to forget the first term and write only eᵗ(-sin t) (choice B), which comes from differentiating just the cosine factor. The product rule requires both parts: keep track of differentiating each factor while holding the other constant.

This problem requires the product rule to differentiate y(t) = (t - 1)tan t. The product rule states that if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = t - 1 with u'(t) = 1, and v(t) = tan t with v'(t) = sec²t. Applying the product rule: dy/dt = (1)(tan t) + (t - 1)(sec²t) = tan t + (t - 1)sec²t. A common error is to write only (t - 1)sec²t (choice B), forgetting the term that comes from differentiating the first factor. Remember the product rule pattern: derivative of first times second, plus first times derivative of second—both terms are essential.

For $D(p)=(10-p)(p^2)$, we apply the product rule with $u=10-p$ and $v=p^2$. We have $u'=-1$ and $v'=2p$. Using the product rule: $D'(p)=(-1)(p^2)+(10-p)(2p)=-p^2+(10-p)2p$. Choice B shows only $(10-p)2p$, which is just the second term and misses the $-p^2$ from the first term. The product rule formula $(uv)'=u'v+uv'$ requires both terms; omitting either one gives an incomplete derivative.

This problem requires the product rule to find the derivative of a model given as a product of two functions. To apply the product rule, identify u(x) = x + 1 with u'(x) = 1, and v(x) = ln(x²) with v'(x) = 2/x. Then, the derivative M'(x) is u'(x)v(x) + u(x)v'(x), which expands to ln(x²) + (x + 1)(2/x). This matches choice A, providing the complete derivative. A tempting distractor like choice B omits the logarithmic term, failing to account for the derivative of the linear function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.

This problem requires the product rule to find the derivative of the temperature model given as a product. Let u(t) = √t with u'(t) = 1/(2√t), and v(t) = t² + 1 with v'(t) = 2t. Then, T'(t) = u'(t)v(t) + u(t)v'(t) = [1/(2√t)](t² + 1) + √t (2t), which is choice A. This expression fully captures the differentiation of both factors. Choice D fails as a distractor by omitting the first term and only differentiating the second factor. A transferable strategy is to rewrite functions in exponential form if needed, but always apply the product rule systematically to both parts.

This problem requires the product rule to differentiate the cost function as a product. Set u(x) = $x^2 - 4x$ with u'(x) = $2x - 4$, and v(x) = $\sqrt{x}$ with v'(x) = $\frac{1}{2\sqrt{x}}$. Thus, C'(x) = u'(x)v(x) + u(x)v'(x) = $(2x - 4)\sqrt{x} + (x^2 - 4x)(\frac{1}{2\sqrt{x}})$, matching choice A. This correctly handles the square root derivative. Choice D is a distractor as it omits the second term, only using the polynomial's derivative. A useful strategy is to express roots as exponents to ease differentiation, but always include both product rule terms.