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.

To find $R'(x)$ for $R(x)=(x+1)\ln(x)$, we apply the product rule. The product rule tells us that $(uv)'=u'v+uv'$ where $u=x+1$ and $v=\ln(x)$. We have $u'=1$ and $v'=\frac{1}{x}$. Substituting into the product rule: $R'(x)=(1)(\ln(x))+(x+1)(\frac{1}{x})=\ln(x)+\frac{x+1}{x}$. Choice C shows only the second term $(x+1)\cdot\frac{1}{x}$, missing the crucial first term $\ln(x)$. Always remember to include both parts of the product rule: the derivative of the first function times the second, plus the first function times the derivative of the second.

This problem requires the product rule to differentiate $s(t) = \sin(t)(t^2 - 4)$. The product rule tells us that if $f(t) = u(t) \cdot v(t)$, then $f'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t)$. Here, $u(t) = \sin(t)$ with $u'(t) = \cos(t)$, and $v(t) = t^2 - 4$ with $v'(t) = 2t$. Applying the product rule: $s'(t) = \cos(t) \cdot(t^2 - 4) + \sin(t) \cdot 2t = \cos(t)(t^2 - 4) + \sin(t)(2t)$. Choice C incorrectly uses only the first term, omitting the crucial second part where we multiply $\sin(t)$ by the derivative of $(t^2 - 4)$. The product rule always yields a sum of two products—both parts are essential for the correct derivative.

This problem requires the product rule to differentiate $\Pi(x) = (x^2 + 3x)\ln(2x)$. The product rule states that if $f(x) = u(x) \cdot v(x)$, then $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$. Here, $u(x) = x^2 + 3x$ with $u'(x) = 2x + 3$, and $v(x) = \ln(2x)$ with $v'(x) = \frac{1}{2x} \cdot 2 = \frac{1}{x}$ by the chain rule. Applying the product rule: $\Pi'(x) = (2x + 3) \cdot \ln(2x) + (x^2 + 3x) \cdot \frac{1}{x} = (2x + 3)\ln(2x) + \frac{x^2 + 3x}{x}$. Choice E incorrectly computes the derivative of $\ln(2x)$ as $\frac{1}{2x}$, missing that $\frac{d}{dx}[\ln(2x)] = \frac{1}{2x} \cdot 2 = \frac{1}{x}$. When differentiating logarithms of linear functions, remember to apply the chain rule correctly.

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 P(t) = t⁴√t = t⁴ · t^(1/2). 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) = t⁴ with u'(t) = 4t³, and v(t) = √t = t^(1/2) with v'(t) = (1/2)t^(-1/2) = 1/(2√t). Applying the product rule: P'(t) = (4t³)(√t) + (t⁴)(1/(2√t)) = 4t³√t + t⁴/(2√t). A tempting shortcut is to combine t⁴√t = t^(9/2) and differentiate directly to get (9/2)t^(7/2) (choice E), but the question asks us to use the product rule form. When applying the product rule, maintain the structure of two separate terms rather than simplifying first.

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.