Power, Product & Quotient Rules - Business Calculus

Find the derivative, $\frac{dy}{dx}$ , for the function $y = \frac{2x^2 + x - 1}{x^2 + 3}$ .
Answer: $\frac{-x^2 + 14x + 3}{(x^2+3)^2}$ Use the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ . Let $u(x) = 2x^2 + x - 1$ , so $u'(x) = 4x + 1$ . Let $v(x) = x^2 + 3$ , so $v'(x) = 2x$ . Plugging into the formula: $\frac{dy}{dx} = \frac{(4x+1)(x^2+3) - (2x^2+x-1)(2x)}{(x^2+3)^2}$ Expand the numerator: Numerator $= (4x^3 + 12x + x^2 + 3) - (4x^3 + 2x^2 - 2x)$ Numerator $= 4x^3 + x^2 + 12x + 3 - 4x^3 - 2x^2 + 2x$ Numerator $= -x^2 + 14x + 3$ So, the derivative is $\frac{-x^2 + 14x + 3}{(x^2+3)^2}$ . (B) This results from reversing the terms in the numerator of the quotient rule, calculating $uv' - u'v$ instead of $u'v - uv'$ . (C) This is the result of the common mistake of taking the derivative of the numerator divided by the derivative of the denominator, $u'/v'$ . (D) This results from correctly calculating the numerator but forgetting to square the denominator in the quotient rule.
Let $h(x) = (x^3 + 2x) g(x)$ . Given that $g(2) = -1$ and $g'(2) = 3$ , find the value of $h'(2)$ .
Answer: $22$ To find $h'(x)$ , we use the product rule. Let $f(x) = x^3 + 2x$ . Then $h(x) = f(x)g(x)$ . The product rule states that $h'(x) = f'(x)g(x) + f(x)g'(x)$ . First, we need to find $f(x)$ and its derivative $f'(x) = 3x^2 + 2$ . Next, we evaluate $f(x)$ and $f'(x)$ at $x=2$ : $f(2) = (2)^3 + 2(2) = 8 + 4 = 12$ . $f'(2) = 3(2)^2 + 2 = 3(4) + 2 = 12 + 2 = 14$ . Now, substitute all the given and calculated values into the product rule formula at $x=2$ : $h'(2) = f'(2)g(2) + f(2)g'(2)$ $h'(2) = (14)(-1) + (12)(3)$ $h'(2) = -14 + 36 = 22$ . (B) This is the result of the common misconception that the derivative of a product is the product of the derivatives, $f'(2)g'(2) = (14)(3) = 42$ . (C) This result comes from an incorrect formula, such as $f(2)g(2) + f'(2)g'(2) = (12)(-1) + (14)(3) = -12 + 42 = 30$ . (D) This could result from confusing the product rule with the quotient rule, using a minus sign: $f'(2)g(2) - f(2)g'(2) = (14)(-1) - (12)(3) = -14 - 36 = -50$ .
The price-demand function for a product is $p(x) = \frac{4000}{0.5x + 8}$ , where $x$ is the number of units sold. The revenue is given by $R(x) = x \cdot p(x)$ . Which of the following is the marginal revenue function, $R'(x)$ ?
Answer: $\frac{32000}{(0.5x + 8)^2}$ First, define the revenue function: $R(x) = x \cdot p(x) = x \cdot \frac{4000}{0.5x + 8} = \frac{4000x}{0.5x + 8}$ . To find the marginal revenue, $R'(x)$ , we use the quotient rule with $u(x) = 4000x$ and $v(x) = 0.5x + 8$ . The derivatives are $u'(x) = 4000$ and $v'(x) = 0.5$ . $R'(x) = \frac{u'v - uv'}{v^2} = \frac{(4000)(0.5x+8) - (4000x)(0.5)}{(0.5x+8)^2}$ Now, simplify the numerator: $R'(x) = \frac{2000x + 32000 - 2000x}{(0.5x+8)^2}$ $R'(x) = \frac{32000}{(0.5x+8)^2}$ . (A) This is the derivative of the price function, $p'(x)$ , not the revenue function. This represents the marginal price. (C) This results from incorrectly adding the terms in the numerator of the quotient rule ( $u'v + uv'$ ) instead of subtracting them. (D) This results from correctly finding the numerator but forgetting to square the denominator in the quotient rule formula.
Given the function $h(t) = \frac{4\sqrt{t}}{t^{1/3} + 3}$ , determine the value of $h'(1)$ .
Answer: $\frac{5}{12}$ To find $h'(t)$ , use the quotient rule. First, write the function with exponents: $h(t) = \frac{4t^{1/2}}{t^{1/3} + 3}$ . Let $u(t) = 4t^{1/2}$ and $v(t) = t^{1/3} + 3$ . Then $u'(t) = 4(\frac{1}{2})t^{-1/2} = 2t^{-1/2}$ . And $v'(t) = \frac{1}{3}t^{-2/3}$ . Apply the quotient rule: $h'(t) = \frac{u'v - uv'}{v^2} = \frac{(2t^{-1/2})(t^{1/3}+3) - (4t^{1/2})(\frac{1}{3}t^{-2/3})}{(t^{1/3}+3)^2}$ . Evaluate at $t=1$ : $h'(1) = \frac{(2(1)^{-1/2})(1^{1/3}+3) - (4(1)^{1/2})(\frac{1}{3}(1)^{-2/3})}{(1^{1/3}+3)^2}$ $h'(1) = \frac{(2)(1+3) - (4)(\frac{1}{3})}{(1+3)^2}$ $h'(1) = \frac{8 - \frac{4}{3}}{16} = \frac{\frac{24}{3} - \frac{4}{3}}{16} = \frac{\frac{20}{3}}{16} = \frac{20}{48} = \frac{5}{12}$ . (A) This result comes from the mistake of calculating $u'(1)/v'(1)$ , which would be $2 / (1/3) = 6$ . (B) This results from using a plus sign instead of a minus sign in the numerator of the quotient rule, giving $\frac{8 + 4/3}{16} = \frac{28/3}{16} = \frac{28}{48} = \frac{7}{12}$ . (D) This results from an error in the power rule, treating the derivative of $\sqrt{t}$ as 1 instead of $\frac{1}{2}t^{-1/2}$ . This makes $u'=4$ , leading to $\frac{4(4) - 4(1/3)}{16} = \frac{16 - 4/3}{16} = \frac{44/3}{16} = \frac{44}{48} = \frac{11}{12}$ .
For which value(s) of $x$ does the graph of the function $f(x) = \frac{x^2 + 12}{2x + 1}$ have a horizontal tangent line?
Answer: $x=3$ and $x=-4$ A horizontal tangent line occurs where the derivative of the function is equal to zero. First, we must find the derivative, $f'(x)$ , using the quotient rule. Let $u(x) = x^2 + 12$ and $v(x) = 2x + 1$ . Then $u'(x) = 2x$ and $v'(x) = 2$ . $f'(x) = \frac{u'v - uv'}{v^2} = \frac{(2x)(2x+1) - (x^2+12)(2)}{(2x+1)^2}$ . Set the derivative equal to zero. A fraction is zero only when its numerator is zero. $(2x)(2x+1) - 2(x^2+12) = 0$ $4x^2 + 2x - 2x^2 - 24 = 0$ $2x^2 + 2x - 24 = 0$ Divide the entire equation by 2 to simplify: $x^2 + x - 12 = 0$ Factor the quadratic equation: $(x+4)(x-3) = 0$ The solutions are $x = -4$ and $x = 3$ . At these two x-values, the function has a horizontal tangent line. (A) This is an incomplete answer; it identifies only one of the correct values. (B) This is an incomplete answer; it identifies only one of the correct values. (D) This results from a sign error when factoring the quadratic equation $x^2 + x - 12 = 0$ as $(x-4)(x+3)=0$ .
A company's revenue function is $R(x) = x^3 \cdot \sqrt{2x + 1}$ where $x$ represents the number of units sold (in thousands). What is $R'(x)$ ?
Answer: $\frac{x^2(7x + 3)}{\sqrt{2x + 1}}$ Using the product rule with $u = x^3$ and $v = (2x + 1)^{1/2}$ : $u' = 3x^2$ and $v' = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = (2x + 1)^{-1/2}$ . So $R'(x) = 3x^2(2x + 1)^{1/2} + x^3(2x + 1)^{-1/2} = \frac{3x^2(2x + 1) + x^3}{\sqrt{2x + 1}} = \frac{6x^3 + 3x^2 + x^3}{\sqrt{2x + 1}} = \frac{x^2(7x + 3)}{\sqrt{2x + 1}}$ . Choice B results from incorrectly computing $6x^3 + x^3 = 8x^3$ . Choice C comes from forgetting the coefficient 2 in the chain rule. Choice D results from computational errors in combining terms.
What is the derivative of the function $f(x) = \frac{x^3 - 8}{x^2 + 2x + 4}$ ? Give your answer in simplest form.
Answer: $1$ This problem can be solved by first simplifying the function $f(x)$ . The numerator, $x^3 - 8$ , is a difference of cubes, which factors as $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$ . For $x^3 - 8 = x^3 - 2^3$ , the factorization is $(x-2)(x^2 + 2x + 4)$ . So, the function is $f(x) = \frac{(x-2)(x^2 + 2x + 4)}{x^2 + 2x + 4}$ . The term $(x^2 + 2x + 4)$ cancels out (it is never zero for real $x$ ), leaving $f(x) = x - 2$ . The derivative of this simplified function is $f'(x) = 1$ . (A) This is the correct, unsimplified result of applying the quotient rule to the original function. The numerator and denominator are identical, so the expression simplifies to 1. A student who correctly applies the quotient rule but fails to simplify would arrive at this complex expression. (B) This is the simplified function $f(x)$ , not its derivative $f'(x)$ . (C) This is the result of the common mistake of taking the derivative of the numerator divided by the derivative of the denominator.
Consider the function $f(x) = \frac{x^2 \sqrt{x+3}}{x-1}$ . What is the value of the derivative $f'(-2)$ ?
Answer: $\frac{2}{9}$ This requires using the quotient rule, and the derivative of the numerator will require the product rule. Let the numerator be $u(x) = x^2 (x+3)^{1/2}$ and the denominator be $v(x) = x-1$ . First, find $u'(x)$ using the product rule: $u'(x) = (2x)(x+3)^{1/2} + (x^2)(\frac{1}{2}(x+3)^{-1/2})$ . Now, apply the quotient rule: $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$ . $v'(x) = 1$ . It is easiest to evaluate the pieces at $x=-2$ before substituting into the full formula: $u(-2) = (-2)^2 \sqrt{-2+3} = 4\sqrt{1} = 4$ . $v(-2) = -2-1 = -3$ . $u'(-2) = (2(-2))\sqrt{-2+3} + ((-2)^2)(\frac{1}{2}(-2+3)^{-1/2}) = (-4)(1) + (4)(\frac{1}{2}(1)^{-1/2}) = -4 + 2 = -2$ . $v'(-2) = 1$ . Now substitute these values into the quotient rule formula: $f'(-2) = \frac{u'(-2)v(-2) - u(-2)v'(-2)}{[v(-2)]^2} = \frac{(-2)(-3) - (4)(1)}{(-3)^2} = \frac{6 - 4}{9} = \frac{2}{9}$ . (B) This results from reversing the terms in the numerator of the quotient rule: $\frac{u(-2)v'(-2) - u'(-2)v(-2)}{[v(-2)]^2} = \frac{4 - 6}{9} = -\frac{2}{9}$ . (C) This results from using a plus sign instead of a minus sign in the numerator of the quotient rule: $\frac{6 + 4}{9} = \frac{10}{9}$ . (D) This results from an error in the product rule for $u'(x)$ , such as adding the derivatives instead of applying the rule correctly.
The function $F(x) = (x^2 + 1)^3 \cdot (2x - 3)^{-2}$ represents a financial model. Which of the following expressions is equivalent to $\frac{F'(x)}{F(x)}$ ?
Answer: $\frac{6x}{x^2 + 1} - \frac{4}{2x - 3}$ This question asks for the logarithmic derivative $\frac{F'(x)}{F(x)}$ . Taking the natural log of both sides: $\ln F(x) = 3\ln(x^2 + 1) - 2\ln(2x - 3)$ . Differentiating both sides: $\frac{F'(x)}{F(x)} = 3 \cdot \frac{2x}{x^2 + 1} - 2 \cdot \frac{2}{2x - 3} = \frac{6x}{x^2 + 1} - \frac{4}{2x - 3}$ . Choice B has the wrong sign on the second term. Choice C has incorrect coefficients (3 instead of 6, and 2 instead of 4). Choice D has incorrect coefficients (12 instead of 6, and 8 instead of 4). The logarithmic derivative method is much more efficient than using the product rule directly for this type of problem.
A manufacturing company models its cost function as $C(q) = \frac{q^3}{(q^2 + 4)^2}$ where $q$ is the quantity produced. The marginal cost function $C'(q)$ simplifies to:
Answer: $\frac{q^2(12 - q^2)}{(q^2 + 4)^3}$ Using the quotient rule with $u = q^3$ and $v = (q^2 + 4)^2$ : $u' = 3q^2$ and $v' = 2(q^2 + 4) \cdot 2q = 4q(q^2 + 4)$ . So $C'(q) = \frac{3q^2(q^2 + 4)^2 - q^3 \cdot 4q(q^2 + 4)}{(q^2 + 4)^4} = \frac{q^2(q^2 + 4)[3(q^2 + 4) - 4q^2]}{(q^2 + 4)^4} = \frac{q^2(3q^2 + 12 - 4q^2)}{(q^2 + 4)^3} = \frac{q^2(12 - q^2)}{(q^2 + 4)^3}$ . Choice B results from miscalculating $3(4) = 8$ instead of 12. Choice C comes from incorrectly using $4 \cdot 4 = 16$ . Choice D results from arithmetic errors in the simplification.
The demand for a product is given by $p(q) = (q^2 + 100)(1 + \frac{200}{q})$ for $q > 0$ . The revenue is $R(q) = q \cdot p(q)$ . Find the marginal revenue at a production level of $q=10$ .
Answer: $4,400$ First, find the revenue function $R(q) = q \cdot p(q)$ . $R(q) = q(q^2 + 100)(1 + \frac{200}{q})$ . It's easiest to simplify this expression before differentiating: $R(q) = (q^3 + 100q)(1 + \frac{200}{q})$ $R(q) = (q^3 + 100q)(1 + 200q^{-1})$ $R(q) = q^3(1+200q^{-1}) + 100q(1+200q^{-1})$ $R(q) = q^3 + 200q^2 + 100q + 20000$ . Now, find the marginal revenue, $R'(q)$ , by differentiating this polynomial: $R'(q) = 3q^2 + 400q + 100$ . Finally, evaluate at $q=10$ : $R'(10) = 3(10)^2 + 400(10) + 100$ $R'(10) = 3(100) + 4000 + 100 = 300 + 4000 + 100 = 4400$ . (A) This is the value of the marginal price, $p'(10)$ , not the marginal revenue. (B) This result is obtained from a sign error when differentiating $200/q$ , treating its derivative as $200q^{-2}$ instead of $-200q^{-2}$ when applying the product rule to the unsimplified form. (C) This is the total revenue $R(10)$ , not the marginal revenue $R'(10)$ .
Two differentiable functions, $f(x)$ and $g(x)$ , have the following properties: $f(3)=4$ , $f'(3)=-2$ , $g(3)=8$ , and $g'(3)=6$ . If $h(x) = \frac{f(x)}{g(x)}$ , what is the value of $h'(3)$ ?
Answer: $-\frac{5}{8}$ To find $h'(x)$ , we apply the quotient rule: $h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ . We need to evaluate this derivative at $x=3$ using the given values: $h'(3) = \frac{f'(3)g(3) - f(3)g'(3)}{[g(3)]^2}$ Substitute the given values into the formula: $h'(3) = \frac{(-2)(8) - (4)(6)}{(8)^2}$ $h'(3) = \frac{-16 - 24}{64}$ $h'(3) = \frac{-40}{64}$ Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 8: $h'(3) = -\frac{5}{8}$ . (A) This is the result of incorrectly applying the product rule formula to the numerator, $f'(3)g(3) + f(3)g'(3) = (-2)(8) + (4)(6) = -16+24=8$ , and then dividing by the denominator, $8/64 = 1/8$ . (B) This results from forgetting to square the denominator in the quotient rule: $\frac{-40}{8} = -5$ . (C) This is the result of reversing the terms in the numerator of the quotient rule: $\frac{f(x)g'(x) - f'(x)g(x)}{[g(x)]^2} = \frac{(4)(6) - (-2)(8)}{64} = \frac{24+16}{64} = \frac{40}{64} = \frac{5}{8}$ .
What is the y-intercept of the line tangent to the graph of $f(x) = \frac{x-3}{x+1}$ at the point where $x=1$ ?
Answer: $-2$ This is a multi-step problem. First, find the point of tangency. At $x=1$ , the y-coordinate is $f(1) = \frac{1-3}{1+1} = \frac{-2}{2} = -1$ . The point is $(1, -1)$ . Next, find the slope of the tangent line by finding the derivative $f'(x)$ using the quotient rule. Let $u=x-3$ and $v=x+1$ . Then $u'=1$ and $v'=1$ . $f'(x) = \frac{u'v - uv'}{v^2} = \frac{(1)(x+1) - (x-3)(1)}{(x+1)^2} = \frac{x+1 - x+3}{(x+1)^2} = \frac{4}{(x+1)^2}$ . Evaluate the slope at $x=1$ : $m = f'(1) = \frac{4}{(1+1)^2} = \frac{4}{4} = 1$ . Now, use the point-slope form of a line, $y - y_1 = m(x - x_1)$ , with point $(1, -1)$ and slope $m=1$ . $y - (-1) = 1(x - 1)$ $y + 1 = x - 1$ $y = x - 2$ . The y-intercept is the value of $y$ when $x=0$ . In this case, $y = 0 - 2 = -2$ . (A) This is the y-coordinate of the point of tangency, not the y-intercept of the tangent line. (B) This is the slope of the tangent line, not the y-intercept. (D) This would be the y-intercept if there were a sign error in the quotient rule calculation, leading to a slope of $m=-1$ and an equation of $y=-x$ .
Let the function $f(x)$ be defined as the product $f(x) = (x^3 - 4x)\sqrt{x}$ . What is the value of $f'(4)$ ?
Answer: $100$ To find the derivative, $f'(x)$ , one can use the product rule. First, rewrite the function as $f(x) = (x^3 - 4x)x^{1/2}$ . Let $u(x) = x^3 - 4x$ and $v(x) = x^{1/2}$ . Then $u'(x) = 3x^2 - 4$ and $v'(x) = \frac{1}{2}x^{-1/2}$ . By the product rule, $f'(x) = u'(x)v(x) + u(x)v'(x) = (3x^2 - 4)x^{1/2} + (x^3 - 4x)(\frac{1}{2}x^{-1/2})$ . Now, evaluate $f'(x)$ at $x=4$ : $f'(4) = (3(4^2) - 4)\sqrt{4} + (4^3 - 4(4))(\frac{1}{2}(4)^{-1/2})$ $f'(4) = (3(16) - 4)(2) + (64 - 16)(\frac{1}{2} \cdot \frac{1}{2})$ $f'(4) = (48 - 4)(2) + (48)(\frac{1}{4})$ $f'(4) = 44(2) + 12 = 88 + 12 = 100$ . Alternatively, one could first expand $f(x) = x^{7/2} - 4x^{3/2}$ . Then $f'(x) = \frac{7}{2}x^{5/2} - 4(\frac{3}{2})x^{1/2} = \frac{7}{2}x^{5/2} - 6x^{1/2}$ . At $x=4$ , $f'(4) = \frac{7}{2}(4)^{5/2} - 6(4)^{1/2} = \frac{7}{2}(32) - 6(2) = 7(16) - 12 = 112 - 12 = 100$ . (A) This is the value of the first term, $u'(4)v(4)$ , in the product rule calculation. (B) This result is obtained by incorrectly applying the product rule as $f' = u'v'$ , which would be $(3x^2-4)(\frac{1}{2}x^{-1/2})$ . At $x=4$ , this gives $(44)(1/4) = 11$ . (D) This results from a calculation error, such as subtracting the two terms in the final step instead of adding them ( $88 - 12 = 76$ ).

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