The Product Rule - AP Calculus BC

The temperature is modeled by $T(t)=e^t\cos t$ . What is $T'(t)$ ?
Answer: $e^t\cos t-e^t\sin t$ 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.
A revenue function is $V(x)=\sqrt{x}(x^2+5)$ . What is $V'(x)$ for $x>0$ ?
Answer: $V'(x)=\dfrac{1}{2\sqrt{x}}(x^2+5)+\sqrt{x}(2x)$ This problem requires the product rule to differentiate V(x) = √x(x² + 5). The product rule states that for f(x) = u(x)·v(x), we get f'(x) = u'(x)·v(x) + u(x)·v'(x). Here, u(x) = √x = $x^{1/2}$ with u'(x) = (1/2) $x^{-1/2}$ = 1/(2√x), and v(x) = x² + 5 with v'(x) = 2x. Applying the product rule: V'(x) = [1/(2√x)]·(x² + 5) + √x·(2x). Choice D incorrectly shows only the first term, omitting the second term √x·(2x). The product rule always produces a sum of two terms—never forget to include both parts when differentiating a product.
A cost function is $C(x)=\sqrt{x}(x^2-4x)$ . What is $C'(x)$ for $x>0$ ?
Answer: $C'(x)=\dfrac{1}{2\sqrt{x}}(x^2-4x)+\sqrt{x}(2x-4)$ This problem requires the product rule to differentiate C(x) = √x(x² - 4x). The product rule states that for h(x) = f(x)·g(x), we get h'(x) = f'(x)·g(x) + f(x)·g'(x). Here, f(x) = √x = $x^{1/2}$ with f'(x) = (1/2) $x^{-1/2}$ = 1/(2√x), and g(x) = x² - 4x with g'(x) = 2x - 4. Applying the product rule: C'(x) = (1/(2√x))(x² - 4x) + (√x)(2x - 4). Choice C incorrectly shows only √x(2x - 4), which is just the second term f·g' and omits the first term f'·g. Remember the product rule mantra: differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
A swimmer's distance is $d(t)=(t-1)e^{2t}$ . What is $d'(t)$ ?
Answer: $d'(t)=e^{2t}+2(t-1)e^{2t}$ For $d(t)=(t-1)e^{2t}$ , we apply the product rule with $u=t-1$ and $v=e^{2t}$ . We find $u'=1$ and $v'=2e^{2t}$ (using the chain rule). By the product rule: $d'(t)=(1)(e^{2t})+(t-1)(2e^{2t})=e^{2t}+2(t-1)e^{2t}$ . Choice C shows only $(t-1)2e^{2t}$ , missing the first term $e^{2t}$ from the product rule. When using the product rule, always include both terms: $(uv)'=u'v+uv'$ , ensuring you don't drop either part of the sum.
A cost function is $C(x)=(\sqrt{x})(x^2+5)$ . What is $C'(x)$ for $x>0$ ?
Answer: $\dfrac{1}{2\sqrt{x}}(x^2+5)+\sqrt{x}(2x)$ This problem requires the product rule to find the derivative of a cost function given as a product of two functions. To apply the product rule, identify u(x) = √x with u'(x) = 1/(2√x), and v(x) = x² + 5 with v'(x) = 2x. Then, the derivative C'(x) is u'(x)v(x) + u(x)v'(x), which expands to [1/(2√x)](x² + 5) + √x (2x). This matches choice C, providing the complete derivative. A tempting distractor like choice A omits the second term, failing to include the derivative of the quadratic function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.
A particle's position is $s(t)=(t^2+1)\sin t$ meters. What is $s'(t)$ ?
Answer: $2t\sin t+(t^2+1)\cos t$ This problem requires the product rule to differentiate s(t) = (t² + 1)sin 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) = t² + 1 with u'(t) = 2t, and v(t) = sin t with v'(t) = cos t. Applying the product rule: s'(t) = (2t)(sin t) + (t² + 1)(cos t) = 2t sin t + (t² + 1)cos t. A common error is differentiating only one part, like finding just 2t sin t (choice A), which forgets to differentiate the second factor. Remember: when differentiating a product, differentiate the first times the second, plus the first times the derivative of the second.
A wave is $y(x)=\sin x\,(1+x^4)$ . What is $y'(x)$ ?
Answer: $y'(x)=\cos x\,(1+x^4)+\sin x\,(4x^3)$ This problem requires the product rule to differentiate y(x) = sin x·(1 + x⁴). 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) = sin x with u'(x) = cos x, and v(x) = 1 + x⁴ with v'(x) = 4x³. Applying the product rule: y'(x) = cos x·(1 + x⁴) + sin x·(4x³). Choice E incorrectly reverses the pattern, showing sin x·(1 + x⁴) + cos x·(4x³), which mixes up which function gets differentiated in each term. The product rule pattern is fixed: derivative of first times second, plus first times derivative of second—never reverse this order.
A signal is modeled by $y(x)=e^{2x}(x^2-x)$ . What is $y'(x)$ ?
Answer: $y'(x)=2e^{2x}(x^2-x)+e^{2x}(2x-1)$ This problem requires the product rule to differentiate y(x) = $e^{2x}$ (x² - x). The product rule states that (f·g)' = f'·g + f·g' for differentiable functions. Here, f(x) = $e^{2x}$ with f'(x) = $2e^{2x}$ by the chain rule, and g(x) = x² - x with g'(x) = 2x - 1. Applying the product rule: y'(x) = ( $2e^{2x}$ )(x² - x) + ( $e^{2x}$ )(2x - 1) = $2e^{2x}$ (x² - x) + $e^{2x}$ (2x - 1). Choice C shows only $2e^{2x}$ (2x - 1), which incorrectly combines terms rather than keeping them separate. The product rule always produces a sum of two distinct terms: f'·g + f·g'.
The height is $h(t)=\sec t\,(t^2-1)$ . What is $h'(t)$ ?
Answer: $h'(t)=\sec t\tan t\,(t^2-1)+\sec t\,(2t)$ This problem requires the product rule to differentiate h(t) = sec t·(t² - 1). The product rule states that for f(x) = u(x)·v(x), we get f'(x) = u'(x)·v(x) + u(x)·v'(x). Here, u(t) = sec t with u'(t) = sec t tan t, and v(t) = t² - 1 with v'(t) = 2t. Applying the product rule: h'(t) = (sec t tan t)·(t² - 1) + sec t·(2t). Choice E incorrectly uses tan t instead of sec t tan t for the derivative of sec t, showing a common confusion between trigonometric derivatives. When using the product rule with trigonometric functions, double-check each derivative—sec t differentiates to sec t tan t, not just tan t.
Let $g(x)=(x^3+2)(x^{-2})$ . What is $g'(x)$ for $x\ne0$ ?
Answer: $g'(x)=3x^2x^{-2}+(x^3+2)(-2x^{-3})$ This problem requires the product rule to differentiate g(x) as a product of two power functions. Identify u(x) = x³ + 2 with u'(x) = 3x², and v(x) = x⁻² with v'(x) = -2x⁻³. Therefore, g'(x) = u'(x)v(x) + u(x)v'(x) = 3x² x⁻² + (x³ + 2)(-2x⁻³), corresponding to choice B. This includes the complete application without missing terms. Choice C is a distractor because it ignores the second term, only differentiating the first factor. To master the product rule, practice identifying negative exponents and ensuring both derivatives are correctly computed and added.
A cost function is $C(x)=(x^2+3x)\ln x$ for $x>0$ . What is $C'(x)$ ?
Answer: $C'(x)=(2x+3)\ln x+(x^2+3x)\cdot\dfrac{1}{x}$ This problem requires the product rule to differentiate C(x) = (x² + 3x)ln x. 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) = x² + 3x with u'(x) = 2x + 3, and v(x) = ln x with v'(x) = 1/x. Applying the product rule: C'(x) = (2x + 3)·ln x + (x² + 3x)·(1/x). Choice B shows only the first term (2x + 3)ln x, omitting the second term where we multiply (x² + 3x) by 1/x. The product rule requires both terms—always include the "first times derivative of second" term even when it involves fractions.
A tank's outflow rate is $R(t)=(t^2+1)e^{-t}$ liters/min. What is $R'(t)$ ?
Answer: $R'(t)=2t e^{-t}-(t^2+1)e^{-t}$ This problem requires the product rule to differentiate R(t) = (t² + 1) $e^{-t}$ . 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(t) = t² + 1 with u'(t) = 2t, and v(t) = $e^{-t}$ with v'(t) = - $e^{-t}$ . Applying the product rule: R'(t) = 2t· $e^{-t}$ + (t² + 1)·(- $e^{-t}$ ) = $2te^{-t}$ - (t² + 1) $e^{-t}$ . Choice B incorrectly computes only the first term u'(t)·v(t), forgetting to add the second term. Remember: the product rule always gives two terms—differentiate the first times the second, plus the first times the derivative of the second.
Let $p(t)=(\tan t)(t^2-1)$ . What is $p'(t)$ ?
Answer: $p'(t)=\sec^2 t\,(t^2-1)+\tan t\,(2t)$ This problem requires the product rule to find p'(t) for a product of tangent and a quadratic. Let u(t) = tan(t) with u'(t) = sec²(t), and v(t) = t² - 1 with v'(t) = 2t. Then, p'(t) = u'(t)v(t) + u(t)v'(t) = sec²(t)(t² - 1) + tan(t)(2t), which is choice B. This incorporates the trigonometric derivative properly. Choice C fails as a distractor by neglecting the second term and the (t² - 1) factor. When dealing with trig functions in products, recall their derivatives accurately and apply the rule to both parts consistently.
If $h(x)=(2x+5)\cos x$ , what is $h'(x)$ ?
Answer: $h'(x)=2\cos x-(2x+5)\sin x$ This problem requires the product rule to find h'(x) for a product involving a linear and cosine function. Let u(x) = 2x + 5 with u'(x) = 2, and v(x) = cos(x) with v'(x) = -sin(x). So, h'(x) = u'(x)v(x) + u(x)v'(x) = 2 cos(x) + (2x + 5)(-sin(x)) = 2 cos(x) - (2x + 5) sin(x), which is choice A. Note the sign from the derivative of cosine. Choice C tempts as a distractor by forgetting the first term and the positive sign. Always double-check the signs in trigonometric derivatives when applying the product rule for accuracy.
A population model is $P(t)=t^4\sqrt{t}$ . What is $P'(t)$ for $t>0$ ?
Answer: $4t^3\sqrt{t}+t^4\cdot \frac{1}{2\sqrt{t}}$ 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.
The revenue is $R(x)=(x+1)\ln(x)$ . What is $R'(x)$ for $x>0$ ?
Answer: $R'(x)=\ln(x)+\dfrac{x+1}{x}$ 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.
A motion model is $v(t)=(t^2-1)\sin(5t)$ . What is $v'(t)$ ?
Answer: $v'(t)=2t\sin(5t)+5(t^2-1)\cos(5t)$ This problem requires the product rule to differentiate v(t) = (t² - 1)sin(5t). The product rule tells us that for f(t) = u(t)·v(t), we get f'(t) = u'(t)·v(t) + u(t)·v'(t). Here, u(t) = t² - 1 with u'(t) = 2t, and v(t) = sin(5t) with v'(t) = 5cos(5t) by the chain rule. Applying the product rule: v'(t) = 2t·sin(5t) + (t² - 1)·5cos(5t) = 2t sin(5t) + 5(t² - 1)cos(5t). Choice B shows only the second term, omitting the product of u'(t) and v(t). The product rule requires both products to be computed and summed—never forget either component.
A model is $M(x)=(x+1)\ln(x^2)$ . What is $M'(x)$ for $x\ne 0$ ?
Answer: $\ln(x^2)+\dfrac{2(x+1)}{x}$ 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.
A growth function is $g(x)=\arctan(x)\,(x^3-2)$ . What is $g'(x)$ ?
Answer: $g'(x)=\frac{x^3-2}{1+x^2}+3x^2\arctan(x)$ 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.
The area is modeled by $A(x)=\sin x\,(x^3-1)$ . What is $A'(x)$ ?
Answer: $\cos x\,(x^3-1)+\sin x\,(3x^2)$ 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.

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