Calculus 1 Quiz: Quotient Rule
20 questions · exam conditions
0:00
Quotient RuleQuestion 1 of 20

Let f(x)=x2+1f(x) = x^2+1 and g(x)=x−1g(x) = x-1. If h(x)=f(x)g(f(x))h(x) = \frac{f(x)}{g(f(x))}, what is h′(1)h'(1)?

−2-2
22
00
−1-1
← Back to quizzes

Calculus 1 Quiz

Calculus 1 Quiz: Quotient Rule

Practice Quotient Rule in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Quotient Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let f(x)=x2+1f(x) = x^2+1 and g(x)=x−1g(x) = x-1. If h(x)=f(x)g(f(x))h(x) = \frac{f(x)}{g(f(x))}, what is h′(1)h'(1)?

  1. −2-2 (correct answer)
  2. 22
  3. 00
  4. −1-1
Explanation: First, find an explicit expression for h(x)h(x). We have f(x)=x2+1f(x) = x^2+1. The composition g(f(x))g(f(x)) is g(x2+1)=(x2+1)−1=x2g(x^2+1) = (x^2+1)-1 = x^2. So, h(x)=x2+1x2h(x) = \frac{x^2+1}{x^2}. Applying the quotient rule: h′(x)=(2x)(x2)−(x2+1)(2x)(x2)2=2x3−2x3−2xx4=−2xx4=−2x3h'(x) = \frac{(2x)(x^2) - (x^2+1)(2x)}{(x^2)^2} = \frac{2x^3 - 2x^3 - 2x}{x^4} = \frac{-2x}{x^4} = -\frac{2}{x^3}. Evaluating at x=1x=1 gives h′(1)=−213=−2h'(1) = -\frac{2}{1^3} = -2. Alternatively, one can simplify h(x)=1+x−2h(x) = 1 + x^{-2}, from which h′(x)=−2x−3h'(x) = -2x^{-3} and h′(1)=−2h'(1)=-2.

Question 2

Let differentiable functions ff and gg be given such that f(2)=3f(2) = 3, f′(2)=−1f'(2) = -1, g(2)=5g(2) = 5, and g′(2)=2g'(2) = 2. If H(x)=g(x)f(x)H(x) = \frac{g(x)}{f(x)}, what is the value of H′(2)H'(2)?

  1. 119\frac{11}{9} (correct answer)
  2. −119-\frac{11}{9}
  3. −1125-\frac{11}{25}
  4. 1125\frac{11}{25}
Explanation: To find H′(x)H'(x), we apply the quotient rule: H′(x)=g′(x)f(x)−g(x)f′(x)[f(x)]2H'(x) = \frac{g'(x)f(x) - g(x)f'(x)}{[f(x)]^2}. Substituting the given values at x=2x=2: H′(2)=g′(2)f(2)−g(2)f′(2)[f(2)]2=(2)(3)−(5)(−1)32=6+59=119H'(2) = \frac{g'(2)f(2) - g(2)f'(2)}{[f(2)]^2} = \frac{(2)(3) - (5)(-1)}{3^2} = \frac{6 + 5}{9} = \frac{11}{9}.

Question 3

Let g(x)=lim⁡h→0cos⁡(x+h)−cos⁡(x)hg(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h}. What is the derivative of the function f(x)=g(x)xf(x) = \frac{g(x)}{x}?

  1. sin⁡(x)−xcos⁡(x)x2\frac{\sin(x) - x\cos(x)}{x^2} (correct answer)
  2. −xsin⁡(x)−cos⁡(x)x2\frac{-x\sin(x) - \cos(x)}{x^2}
  3. xcos⁡(x)−sin⁡(x)x2\frac{x\cos(x) - \sin(x)}{x^2}
  4. sin⁡(x)+xcos⁡(x)x2\frac{\sin(x) + x\cos(x)}{x^2}
Explanation: First, recognize that g(x)g(x) is the definition of the derivative of cos⁡(x)\cos(x). Therefore, g(x)=−sin⁡(x)g(x) = -\sin(x). The problem asks for the derivative of f(x)=−sin⁡(x)xf(x) = \frac{-\sin(x)}{x}. Using the quotient rule: f′(x)=(−cos⁡(x))(x)−(−sin⁡(x))(1)x2=−xcos⁡(x)+sin⁡(x)x2=sin⁡(x)−xcos⁡(x)x2f'(x) = \frac{(-\cos(x))(x) - (-\sin(x))(1)}{x^2} = \frac{-x\cos(x) + \sin(x)}{x^2} = \frac{\sin(x) - x\cos(x)}{x^2}.

Question 4

Given that y=cos⁡(x)1−sin⁡(x)y = \frac{\cos(x)}{1-\sin(x)}, which of the following is an expression for dydx\frac{dy}{dx}?

  1. 11−sin⁡(x)\frac{1}{1-\sin(x)} (correct answer)
  2. −sin⁡(x)−cos⁡(x)\frac{-\sin(x)}{-\cos(x)}
  3. sin⁡(x)−1(1−sin⁡(x))2\frac{\sin(x)-1}{(1-\sin(x))^2}
  4. cos⁡2(x)−sin⁡(x)(1−sin⁡(x))2\frac{\cos^2(x)-\sin(x)}{(1-\sin(x))^2}
Explanation: Using the quotient rule: dydx=(−sin⁡x)(1−sin⁡x)−(cos⁡x)(−cos⁡x)(1−sin⁡x)2\frac{dy}{dx} = \frac{(-\sin x)(1-\sin x) - (\cos x)(-\cos x)}{(1-\sin x)^2}. Expanding the numerator gives −sin⁡x+sin⁡2x+cos⁡2x-\sin x + \sin^2 x + \cos^2 x. Using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, the numerator becomes 1−sin⁡x1 - \sin x. So, dydx=1−sin⁡x(1−sin⁡x)2=11−sin⁡x\frac{dy}{dx} = \frac{1-\sin x}{(1-\sin x)^2} = \frac{1}{1-\sin x}.

Question 5

Let f(x)=x2−kx+kf(x) = \frac{x^2-k}{x+k}, where kk is a non-zero constant. The tangent line to the graph of ff at x=0x=0 is parallel to the line y=−x+5y = -x+5. What is the value of kk?

  1. 11
  2. −1-1 (correct answer)
  3. 22
  4. −2-2
Explanation: The slope of the line y=−x+5y=-x+5 is -1. For the tangent line to be parallel, its slope must also be -1. The slope of the tangent line is given by the derivative f′(x)f'(x). Using the quotient rule: f′(x)=(2x)(x+k)−(x2−k)(1)(x+k)2=2x2+2kx−x2+k(x+k)2=x2+2kx+k(x+k)2f'(x) = \frac{(2x)(x+k) - (x^2-k)(1)}{(x+k)^2} = \frac{2x^2+2kx-x^2+k}{(x+k)^2} = \frac{x^2+2kx+k}{(x+k)^2}. We need the slope at x=0x=0 to be -1. So, we evaluate f′(0)f'(0): f′(0)=02+2k(0)+k(0+k)2=kk2=1kf'(0) = \frac{0^2+2k(0)+k}{(0+k)^2} = \frac{k}{k^2} = \frac{1}{k}. Setting the slope equal to -1: 1k=−1\frac{1}{k} = -1, which implies k=−1k = -1.

Question 6

What is the equation of the line tangent to the graph of f(x)=3x+2x+1f(x) = \frac{3x+2}{x+1} at x=1x=1?

  1. y=14x+94y = \frac{1}{4}x + \frac{9}{4} (correct answer)
  2. y=−14x+114y = -\frac{1}{4}x + \frac{11}{4}
  3. y=12x+2y = \frac{1}{2}x + 2
  4. y=−4x+132y = -4x + \frac{13}{2}
Explanation: First, find the point of tangency: f(1)=3(1)+21+1=52f(1) = \frac{3(1)+2}{1+1} = \frac{5}{2}. The point is (1,52)(1, \frac{5}{2}). Next, find the slope by calculating the derivative using the quotient rule: f′(x)=3(x+1)−(3x+2)(1)(x+1)2=3x+3−3x−2(x+1)2=1(x+1)2f'(x) = \frac{3(x+1) - (3x+2)(1)}{(x+1)^2} = \frac{3x+3-3x-2}{(x+1)^2} = \frac{1}{(x+1)^2}. The slope at x=1x=1 is f′(1)=1(1+1)2=14f'(1) = \frac{1}{(1+1)^2} = \frac{1}{4}. Using the point-slope form: y−52=14(x−1)y - \frac{5}{2} = \frac{1}{4}(x-1), which simplifies to y=14x−14+104=14x+94y = \frac{1}{4}x - \frac{1}{4} + \frac{10}{4} = \frac{1}{4}x + \frac{9}{4}.

Question 7

If f(x)=e2xx2f(x) = \frac{e^{2x}}{x^2}, then f′(x)=f'(x) =?

  1. 2e2x2x\frac{2e^{2x}}{2x}
  2. e2x(x−1)x3\frac{e^{2x}(x-1)}{x^3}
  3. 2e2x(x−2)x3\frac{2e^{2x}(x-2)}{x^3}
  4. 2e2x(x−1)x3\frac{2e^{2x}(x-1)}{x^3} (correct answer)
Explanation: When you encounter a function that's a quotient of two expressions, like f(x)=e2xx2f(x) = \frac{e^{2x}}{x^2}, you need the quotient rule for differentiation: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. Here, u=e2xu = e^{2x} and v=x2v = x^2. First, find the derivatives: u′=2e2xu' = 2e^{2x} (using the chain rule) and v′=2xv' = 2x. Applying the quotient rule: f′(x)=(2e2x)(x2)−(e2x)(2x)(x2)2=2e2xx2−2xe2xx4f'(x) = \frac{(2e^{2x})(x^2) - (e^{2x})(2x)}{(x^2)^2} = \frac{2e^{2x}x^2 - 2xe^{2x}}{x^4} Factor out 2xe2x2xe^{2x} from the numerator: f′(x)=2xe2x(x−1)x4=2e2x(x−1)x3f'(x) = \frac{2xe^{2x}(x - 1)}{x^4} = \frac{2e^{2x}(x-1)}{x^3} This matches choice D. Choice A, 2e2x2x\frac{2e^{2x}}{2x}, appears to treat this as if you could differentiate the numerator and denominator separately—that's not how differentiation works for quotients. Choice B, e2x(x−1)x3\frac{e^{2x}(x-1)}{x^3}, correctly applies the quotient rule structure but misses the factor of 2 that comes from differentiating e2xe^{2x}. Choice C, 2e2x(x−2)x3\frac{2e^{2x}(x-2)}{x^3}, makes an algebraic error when factoring—likely confusing terms in the quotient rule application. Study tip: When using the quotient rule, write out each step clearly and factor carefully at the end. Many errors occur during the algebraic simplification, not in applying the rule itself.

Question 8

Let f(x)=ax+1x2−3f(x) = \frac{ax+1}{x^2-3}, where aa is a constant. For what value of aa is f′(2)=−1f'(2) = -1?

  1. −57-\frac{5}{7}
  2. −37-\frac{3}{7} (correct answer)
  3. 57\frac{5}{7}
  4. −13-\frac{1}{3}
Explanation: First, find the derivative f′(x)f'(x) using the quotient rule: f′(x)=(a)(x2−3)−(ax+1)(2x)(x2−3)2=ax2−3a−2ax2−2x(x2−3)2=−ax2−2x−3a(x2−3)2f'(x) = \frac{(a)(x^2-3) - (ax+1)(2x)}{(x^2-3)^2} = \frac{ax^2 - 3a - 2ax^2 - 2x}{(x^2-3)^2} = \frac{-ax^2 - 2x - 3a}{(x^2-3)^2}. Now, evaluate at x=2x=2: f′(2)=−a(22)−2(2)−3a(22−3)2=−4a−4−3a(1)2=−7a−4f'(2) = \frac{-a(2^2) - 2(2) - 3a}{(2^2-3)^2} = \frac{-4a - 4 - 3a}{(1)^2} = -7a - 4. We are given that f′(2)=−1f'(2) = -1, so we set up the equation: −7a−4=−1-7a - 4 = -1. Solving for aa gives −7a=3-7a = 3, so a=−37a = -\frac{3}{7}.

Question 9

The concentration CC of a medication in a patient's bloodstream tt hours after injection is modeled by C(t)=3tt2+9C(t) = \frac{3t}{t^2+9} for t≥0t \ge 0. At what time tt is the concentration of the medication increasing most rapidly?

  1. t=3t=3
  2. t=33t=3\sqrt{3} (correct answer)
  3. t=9t=9
  4. t=3t=\sqrt{3}
Explanation: The rate of change of concentration is C′(t)C'(t). To find when this rate is at a maximum, we need to find the derivative of C′(t)C'(t), which is C′′(t)C''(t), and find its critical points. First, find C′(t)C'(t) using the Quotient Rule: C′(t)=(t2+9)(3)−(3t)(2t)(t2+9)2=3t2+27−6t2(t2+9)2=27−3t2(t2+9)2C'(t) = \frac{(t^2+9)(3) - (3t)(2t)}{(t^2+9)^2} = \frac{3t^2+27-6t^2}{(t^2+9)^2} = \frac{27-3t^2}{(t^2+9)^2} This is the rate of change. To maximize it, we find its derivative, C′′(t)C''(t), using the Quotient Rule again: C′′(t)=(−6t)(t2+9)2−(27−3t2)(2(t2+9)(2t))((t2+9)2)2C''(t) = \frac{(-6t)(t^2+9)^2 - (27-3t^2)(2(t^2+9)(2t))}{((t^2+9)^2)^2} Set C′′(t)=0C''(t)=0. This occurs when the numerator is 0. We can factor out −6t(t2+9)-6t(t^2+9) to simplify: Numerator = (t2+9)[(−6t)(t2+9)−4t(27−3t2)]=0(t^2+9)[(-6t)(t^2+9) - 4t(27-3t^2)] = 0 (−6t3−54t)−(108t−12t3)=0(-6t^3 - 54t) - (108t - 12t^3) = 0 6t3−162t=06t^3 - 162t = 0 6t(t2−27)=06t(t^2 - 27) = 0 This gives t=0t=0 or t2=27t^2=27. For t>0t>0, we have t=27=33t = \sqrt{27} = 3\sqrt{3}. This corresponds to an inflection point on the graph of C(t)C(t), where the rate of increase is maximal.

Question 10

Let h(x)=xexx2+1h(x) = \frac{x e^x}{x^2+1}. What is the value of h′(1)h'(1)?

  1. 00
  2. ee
  3. −e2-\frac{e}{2}
  4. e2\frac{e}{2} (correct answer)
Explanation: We must use the quotient rule combined with the product rule for the numerator. Let f(x)=xexf(x) = xe^x and g(x)=x2+1g(x) = x^2+1. By the product rule, f′(x)=(1)ex+x(ex)=ex(1+x)f'(x) = (1)e^x + x(e^x) = e^x(1+x). Now, applying the quotient rule: h′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2=ex(1+x)(x2+1)−(xex)(2x)(x2+1)2h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = \frac{e^x(1+x)(x^2+1) - (xe^x)(2x)}{(x^2+1)^2}. Evaluating at x=1x=1: h′(1)=e1(1+1)(12+1)−(1e1)(2(1))(12+1)2=e(2)(2)−2e22=4e−2e4=2e4=e2h'(1) = \frac{e^1(1+1)(1^2+1) - (1e^1)(2(1))}{(1^2+1)^2} = \frac{e(2)(2) - 2e}{2^2} = \frac{4e - 2e}{4} = \frac{2e}{4} = \frac{e}{2}.

Question 11

The derivative of f(x)=tan⁡(x)f(x) = \tan(x) can be derived by writing tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)} and applying the quotient rule. Which of the following is the correct intermediate expression for f′(x)f'(x) before any trigonometric identities are used for simplification?

  1. cos⁡2(x)−sin⁡2(x)cos⁡2(x)\frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)}
  2. sin⁡(x)(−sin⁡(x))−cos⁡(x)cos⁡(x)cos⁡2(x)\frac{\sin(x)(-\sin(x)) - \cos(x)\cos(x)}{\cos^2(x)}
  3. cos⁡(x)cos⁡(x)−sin⁡(x)(−sin⁡(x))cos⁡2(x)\frac{\cos(x)\cos(x) - \sin(x)(-\sin(x))}{\cos^2(x)} (correct answer)
  4. cos⁡(x)cos⁡(x)+sin⁡(x)(−sin⁡(x))sin⁡2(x)\frac{\cos(x)\cos(x) + \sin(x)(-\sin(x))}{\sin^2(x)}
Explanation: Let u(x)=sin⁡(x)u(x) = \sin(x) and v(x)=cos⁡(x)v(x) = \cos(x). Then u′(x)=cos⁡(x)u'(x) = \cos(x) and v′(x)=−sin⁡(x)v'(x) = -\sin(x). The quotient rule is u′v−uv′v2\frac{u'v - uv'}{v^2}. Applying this gives: f′(x)=(cos⁡(x))(cos⁡(x))−(sin⁡(x))(−sin⁡(x))(cos⁡(x))2=cos⁡2(x)+sin⁡2(x)cos⁡2(x)f'(x) = \frac{(\cos(x))(\cos(x)) - (\sin(x))(-\sin(x))}{(\cos(x))^2} = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}. Option C correctly represents this step.

Question 12

Let h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)} and k(x)=g(x)f(x)k(x) = \frac{g(x)}{f(x)}. Assuming f(a)≠0f(a) \neq 0, g(a)≠0g(a) \neq 0, and h′(a)h'(a) exists, which of the following correctly expresses k′(a)k'(a) in terms of h(a)h(a) and h′(a)h'(a)?

  1. k′(a)=−h′(a)k'(a) = -h'(a)
  2. k′(a)=1h′(a)k'(a) = \frac{1}{h'(a)}
  3. k′(a)=−h′(a)[h(a)]2k'(a) = -\frac{h'(a)}{[h(a)]^2} (correct answer)
  4. k′(a)=h′(a)[h(a)]2k'(a) = \frac{h'(a)}{[h(a)]^2}
Explanation: Since k(x)=1h(x)k(x) = \frac{1}{h(x)}, we can find k′(x)k'(x) using the quotient rule (or reciprocal rule). Let u(x)=1u(x)=1 and v(x)=h(x)v(x)=h(x). Then k′(x)=u′v−uv′v2=(0)h(x)−(1)h′(x)[h(x)]2=−h′(x)[h(x)]2k'(x) = \frac{u'v - uv'}{v^2} = \frac{(0)h(x) - (1)h'(x)}{[h(x)]^2} = -\frac{h'(x)}{[h(x)]^2}. Evaluating at x=ax=a gives k′(a)=−h′(a)[h(a)]2k'(a) = -\frac{h'(a)}{[h(a)]^2}.

Question 13

The line y=2x−1y=2x-1 is tangent to the graph of a function f(x)f(x) at x=3x=3. The line y=−x+7y=-x+7 is tangent to the graph of a function g(x)g(x) at x=3x=3.

Given the information in the passage, what is the derivative of h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)} at x=3x=3?

  1. −1316-\frac{13}{16}
  2. 1316\frac{13}{16} (correct answer)
  3. −2-2
  4. 33
Explanation: From the tangent line information, we can deduce the function and derivative values at x=3x=3. For f(x)f(x), the point of tangency is f(3)=2(3)−1=5f(3) = 2(3)-1 = 5 and the slope is f′(3)=2f'(3) = 2. For g(x)g(x), the point of tangency is g(3)=−3+7=4g(3) = -3+7 = 4 and the slope is g′(3)=−1g'(3) = -1. Now, apply the quotient rule to h(x)h(x): h′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}. At x=3x=3, h′(3)=f′(3)g(3)−f(3)g′(3)[g(3)]2=(2)(4)−(5)(−1)42=8+516=1316h'(3) = \frac{f'(3)g(3) - f(3)g'(3)}{[g(3)]^2} = \frac{(2)(4) - (5)(-1)}{4^2} = \frac{8+5}{16} = \frac{13}{16}.

Question 14

Let h(x)=f(x2)xh(x) = \frac{f(x^2)}{x}. Given that f(1)=3f(1) = 3 and f′(1)=−2f'(1) = -2, find h′(1)h'(1).

  1. −7-7 (correct answer)
  2. −5-5
  3. −4-4
  4. −1-1
Explanation: We use the quotient rule and the chain rule. The derivative of the numerator is ddxf(x2)=f′(x2)⋅2x\frac{d}{dx}f(x^2) = f'(x^2) \cdot 2x. Applying the quotient rule: h′(x)=(f′(x2)⋅2x)(x)−(f(x2))(1)x2h'(x) = \frac{(f'(x^2) \cdot 2x)(x) - (f(x^2))(1)}{x^2}. Now, substitute x=1x=1: h′(1)=(f′(12)⋅2(1))(1)−f(12)12=f′(1)⋅2−f(1)1=2f′(1)−f(1)h'(1) = \frac{(f'(1^2) \cdot 2(1))(1) - f(1^2)}{1^2} = \frac{f'(1) \cdot 2 - f(1)}{1} = 2f'(1) - f(1). Using the given values, h′(1)=2(−2)−3=−4−3=−7h'(1) = 2(-2) - 3 = -4 - 3 = -7.

Question 15

If f(x)=xx−3f(x) = \frac{x}{x-3}, then what is the expression for the second derivative, f′′(x)f''(x)?

  1. 6(x−3)3\frac{6}{(x-3)^3} (correct answer)
  2. −6(x−3)3\frac{-6}{(x-3)^3}
  3. 3(x−3)2\frac{3}{(x-3)^2}
  4. 6(x−3)4\frac{6}{(x-3)^4}
Explanation: First, find the first derivative using the quotient rule: f′(x)=(1)(x−3)−(x)(1)(x−3)2=x−3−x(x−3)2=−3(x−3)2f'(x) = \frac{(1)(x-3) - (x)(1)}{(x-3)^2} = \frac{x-3-x}{(x-3)^2} = \frac{-3}{(x-3)^2}. To find the second derivative, we can write f′(x)=−3(x−3)−2f'(x) = -3(x-3)^{-2} and use the power rule and chain rule: f′′(x)=−3(−2)(x−3)−3(1)=6(x−3)−3=6(x−3)3f''(x) = -3(-2)(x-3)^{-3}(1) = 6(x-3)^{-3} = \frac{6}{(x-3)^3}. Alternatively, applying the quotient rule to f′(x)f'(x) gives f′′(x)=(0)(x−3)2−(−3)(2(x−3))(x−3)4=6(x−3)(x−3)4=6(x−3)3f''(x) = \frac{(0)(x-3)^2 - (-3)(2(x-3))}{(x-3)^4} = \frac{6(x-3)}{(x-3)^4} = \frac{6}{(x-3)^3}.

Question 16

If f(x)=sin⁡(πx)x2−1f(x) = \frac{\sin(\pi x)}{x^2-1}, what is f′(0)f'(0)?

  1. 00
  2. π\pi
  3. −π-\pi (correct answer)
  4. Undefined
Explanation: We use the quotient rule and the chain rule: f′(x)=(cos⁡(πx)⋅π)(x2−1)−(sin⁡(πx))(2x)(x2−1)2f'(x) = \frac{(\cos(\pi x) \cdot \pi)(x^2-1) - (\sin(\pi x))(2x)}{(x^2-1)^2}. To find f′(0)f'(0), we substitute x=0x=0 into the expression for the derivative: f′(0)=(cos⁡(0)⋅π)(02−1)−(sin⁡(0))(2⋅0)(02−1)2f'(0) = \frac{(\cos(0) \cdot \pi)(0^2-1) - (\sin(0))(2 \cdot 0)}{(0^2-1)^2}. Since cos⁡(0)=1\cos(0) = 1 and sin⁡(0)=0\sin(0) = 0, this simplifies to: f′(0)=(1⋅π)(−1)−(0)(0)(−1)2=−π1=−πf'(0) = \frac{(1 \cdot \pi)(-1) - (0)(0)}{(-1)^2} = \frac{-\pi}{1} = -\pi.

Question 17

Let f(x)=x3−8x2−4f(x) = \frac{x^3-8}{x^2-4} for x≠2x \neq 2. What is the slope of the line tangent to the graph of ff at x=1x=1?

  1. 73\frac{7}{3}
  2. −59-\frac{5}{9}
  3. 11
  4. 59\frac{5}{9} (correct answer)
Explanation: When you need to find the slope of a tangent line at a specific point, you're looking for the derivative of the function evaluated at that point. The slope of the tangent line to f(x)f(x) at x=1x = 1 is simply f′(1)f'(1). To find f′(x)f'(x) for f(x)=x3−8x2−4f(x) = \frac{x^3-8}{x^2-4}, use the quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. Here, u=x3−8u = x^3-8 and v=x2−4v = x^2-4, so u′=3x2u' = 3x^2 and v′=2xv' = 2x. Applying the quotient rule: f′(x)=(3x2)(x2−4)−(x3−8)(2x)(x2−4)2f'(x) = \frac{(3x^2)(x^2-4) - (x^3-8)(2x)}{(x^2-4)^2} Expanding the numerator: f′(x)=3x4−12x2−2x4+16x(x2−4)2=x4−12x2+16x(x2−4)2f'(x) = \frac{3x^4 - 12x^2 - 2x^4 + 16x}{(x^2-4)^2} = \frac{x^4 - 12x^2 + 16x}{(x^2-4)^2} Now evaluate at x=1x = 1: f′(1)=1−12+16(1−4)2=59f'(1) = \frac{1 - 12 + 16}{(1-4)^2} = \frac{5}{9} This confirms answer D is correct. Answer A (73\frac{7}{3}) likely comes from arithmetic errors in the quotient rule application. Answer B (−59-\frac{5}{9}) gets the magnitude right but has the wrong sign, possibly from sign errors in the calculation. Answer C (11) might result from incorrectly simplifying the original function before differentiating or making computational mistakes. Remember: always double-check your quotient rule algebra, especially the signs. Expanding systematically and combining like terms carefully will help you avoid common arithmetic pitfalls in derivative problems.

Question 18

Let f(x)=xx2+1f(x) = \frac{x}{\sqrt{x^2+1}}. Find f′(x)f'(x).

  1. 2x2+1(x2+1)3/2\frac{2x^2+1}{(x^2+1)^{3/2}}
  2. 1−x2(x2+1)3/2\frac{1-x^2}{(x^2+1)^{3/2}}
  3. 12x2+1\frac{1}{2\sqrt{x^2+1}}
  4. 1(x2+1)3/2\frac{1}{(x^2+1)^{3/2}} (correct answer)
Explanation: When you encounter a function with a fraction where both the numerator and denominator contain the variable, the quotient rule is your primary tool. Here we have f(x)=xx2+1f(x) = \frac{x}{\sqrt{x^2+1}}, so we need to apply the quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. Let u=xu = x and v=x2+1=(x2+1)1/2v = \sqrt{x^2+1} = (x^2+1)^{1/2}. Then u′=1u' = 1. For v′v', we need the chain rule: v′=12(x2+1)−1/2⋅2x=xx2+1v' = \frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+1}}. Applying the quotient rule: f′(x)=1⋅x2+1−x⋅xx2+1(x2+1)2=x2+1−x2x2+1x2+1f'(x) = \frac{1 \cdot \sqrt{x^2+1} - x \cdot \frac{x}{\sqrt{x^2+1}}}{(\sqrt{x^2+1})^2} = \frac{\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}}{x^2+1} To simplify the numerator, multiply the first term by x2+1x2+1\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}: x2+1x2+1−x2x2+1x2+1=1x2+1x2+1=1(x2+1)3/2\frac{\frac{x^2+1}{\sqrt{x^2+1}} - \frac{x^2}{\sqrt{x^2+1}}}{x^2+1} = \frac{\frac{1}{\sqrt{x^2+1}}}{x^2+1} = \frac{1}{(x^2+1)^{3/2}} This confirms answer D is correct. Answer A represents a sign error and incorrect coefficient. Answer B comes from confusing terms in the quotient rule application. Answer C results from incorrectly applying the chain rule to just the denominator without considering the quotient structure. Study tip: Always write out each step of the quotient rule explicitly, especially when the denominator requires the chain rule. Double-check your algebra when combining fractions in the numerator.

Question 19

Let h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}. Suppose f(a)=4f(a)=4, f′(a)=−2f'(a)=-2, g(a)=1g(a)=1, and h′(a)=6h'(a)=6. What is the value of g′(a)g'(a)?

  1. −2-2 (correct answer)
  2. −1-1
  3. 11
  4. 22
Explanation: The Quotient Rule states that h′(x)=g(x)f′(x)−f(x)g′(x)[g(x)]2h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}. We are given values at x=ax=a. Let's substitute them into the formula: h′(a)=g(a)f′(a)−f(a)g′(a)[g(a)]2h'(a) = \frac{g(a)f'(a) - f(a)g'(a)}{[g(a)]^2} Substitute the known values: h′(a)=6h'(a)=6, f(a)=4f(a)=4, f′(a)=−2f'(a)=-2, and g(a)=1g(a)=1. 6=(1)(−2)−(4)g′(a)126 = \frac{(1)(-2) - (4)g'(a)}{1^2} 6=−2−4g′(a)6 = -2 - 4g'(a) Now, we solve for g′(a)g'(a): 8=−4g′(a)8 = -4g'(a) g′(a)=8−4=−2g'(a) = \frac{8}{-4} = -2

Question 20

If g(x)=ln⁡(x)x3g(x) = \frac{\ln(x)}{x^3}, what is g′(x)g'(x)?

  1. 1−3ln⁡(x)x4\frac{1 - 3\ln(x)}{x^4} (correct answer)
  2. 3ln⁡(x)−1x4\frac{3\ln(x) - 1}{x^4}
  3. 1−3ln⁡(x)x6\frac{1 - 3\ln(x)}{x^6}
  4. 13x3\frac{1}{3x^3}
Explanation: Using the quotient rule, g′(x)=(ddxln⁡(x))⋅x3−ln⁡(x)⋅(ddxx3)(x3)2g'(x) = \frac{(\frac{d}{dx}\ln(x)) \cdot x^3 - \ln(x) \cdot (\frac{d}{dx}x^3)}{(x^3)^2}. The derivatives are ddxln⁡(x)=1x\frac{d}{dx}\ln(x) = \frac{1}{x} and ddxx3=3x2\frac{d}{dx}x^3 = 3x^2. Substituting these in gives: g′(x)=(1x)(x3)−(ln⁡x)(3x2)x6=x2−3x2ln⁡(x)x6g'(x) = \frac{(\frac{1}{x})(x^3) - (\ln x)(3x^2)}{x^6} = \frac{x^2 - 3x^2\ln(x)}{x^6}. Factoring out x2x^2 from the numerator and simplifying yields g′(x)=x2(1−3ln⁡(x))x6=1−3ln⁡(x)x4g'(x) = \frac{x^2(1 - 3\ln(x))}{x^6} = \frac{1 - 3\ln(x)}{x^4}.