Calculus 1 Quiz: Derivative Definition And Notation
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Derivative Definition And NotationQuestion 1 of 20

Let ff be a function differentiable at x=ax=a. The expression lim⁡h→0f(a+h)−f(a−h)2h\lim_{h \to 0} \frac{f(a+h) - f(a-h)}{2h} is known as the symmetric difference quotient. If f′(a)f'(a) exists, what is the value of this limit?

2f′(a)2f'(a)
12f′(a)\frac{1}{2}f'(a)
f′(a)f'(a)
00
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Calculus 1 Quiz

Calculus 1 Quiz: Derivative Definition And Notation

Practice Derivative Definition And Notation 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 Derivative Definition And Notation, 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.

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Question 1

Let ff be a function differentiable at x=ax=a. The expression lim⁡h→0f(a+h)−f(a−h)2h\lim_{h \to 0} \frac{f(a+h) - f(a-h)}{2h} is known as the symmetric difference quotient. If f′(a)f'(a) exists, what is the value of this limit?

  1. 2f′(a)2f'(a)
  2. 12f′(a)\frac{1}{2}f'(a)
  3. f′(a)f'(a) (correct answer)
  4. 00
Explanation: We can rewrite the expression by adding and subtracting f(a)f(a) in the numerator: lim⁡h→0f(a+h)−f(a)−f(a−h)+f(a)2h\lim_{h \to 0} \frac{f(a+h) - f(a) - f(a-h) + f(a)}{2h} =lim⁡h→0[f(a+h)−f(a)2h+f(a)−f(a−h)2h]= \lim_{h \to 0} \left[ \frac{f(a+h) - f(a)}{2h} + \frac{f(a) - f(a-h)}{2h} \right]. This can be split into two limits: 12lim⁡h→0f(a+h)−f(a)h+12lim⁡h→0f(a−h)−f(a)−h\frac{1}{2} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} + \frac{1}{2} \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h}. Both of these limits are equal to f′(a)f'(a). Therefore, the expression simplifies to 12f′(a)+12f′(a)=f′(a)\frac{1}{2}f'(a) + \frac{1}{2}f'(a) = f'(a).

Question 2

The expression lim⁡h→0e2(x+h)−e2xh\lim_{h \to 0} \frac{e^{2(x+h)} - e^{2x}}{h} represents the derivative of which function?

  1. f(x)=e2xf(x) = e^{2x} (correct answer)
  2. f(x)=exf(x) = e^x
  3. f(x)=e2(x+h)f(x) = e^{2(x+h)}
  4. f(x)=ex+hf(x) = e^{x+h}
Explanation: The definition of the derivative is f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. By comparing the given expression with this definition, we can identify f(x)=e2xf(x) = e^{2x}. When we replace xx with x+hx+h in this function, we get f(x+h)=e2(x+h)f(x+h) = e^{2(x+h)}, which matches the first term in the numerator.

Question 3

Let ff be a differentiable function. Which of the following is the definition of f′′(x)f''(x), the second derivative of f(x)f(x)?

  1. lim⁡h→0f(x+2h)−f(x)2h\lim_{h \to 0} \frac{f(x+2h) - f(x)}{2h}
  2. lim⁡h→0f(x+h)−2f(x)+f(x−h)h2\lim_{h \to 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}
  3. (lim⁡h→0f(x+h)−f(x)h)2\left( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \right)^2
  4. lim⁡h→0f′(x+h)−f′(x)h\lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h} (correct answer)
Explanation: When you encounter questions about higher-order derivatives, remember that each derivative is built from the previous one using the fundamental limit definition. The second derivative is simply the derivative of the first derivative. The correct answer is D because f′′(x)f''(x) is defined as the derivative of f′(x)f'(x). Since the derivative of any function g(x)g(x) is lim⁡h→0g(x+h)−g(x)h\lim_{h \to 0} \frac{g(x+h) - g(x)}{h}, we substitute g(x)=f′(x)g(x) = f'(x) to get f′′(x)=lim⁡h→0f′(x+h)−f′(x)hf''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h}. This follows directly from applying the definition of derivative to the first derivative function. Option A gives you a difference quotient with step size 2h2h, which would approximate 2f′(x)2f'(x), not the second derivative. Option B represents a second-order finite difference formula that can approximate the second derivative numerically, but it's not the formal definition—it's a computational shortcut used in numerical analysis. Option C is a common trap: it's the square of the first derivative, (f′(x))2(f'(x))^2, not the second derivative. The notation similarity between "second derivative" and "derivative squared" causes confusion here. Remember this pattern: to find the nn-th derivative's definition, apply the basic derivative limit formula to the (n−1)(n-1)-th derivative. The second derivative is always "the derivative of the derivative," so start with f′(x)f'(x) and apply the limit definition once more.

Question 4

If f(x)=xf(x) = \sqrt{x}, which expression correctly represents dfdx\frac{df}{dx} using the alternate limit definition of the derivative?

  1. lim⁡z→xz−xz−x\lim_{z \to x} \frac{\sqrt{z} - \sqrt{x}}{z-x} (correct answer)
  2. lim⁡x→ax−ax\lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x}
  3. lim⁡z→xx−zz−x\lim_{z \to x} \frac{\sqrt{x} - \sqrt{z}}{z-x}
  4. lim⁡z→xz−xz−x\lim_{z \to x} \frac{\sqrt{z} - x}{z-x}
Explanation: The alternate definition of the derivative of f(x)f(x) is f′(x)=lim⁡z→xf(z)−f(x)z−xf'(x) = \lim_{z \to x} \frac{f(z) - f(x)}{z-x}. In this definition, zz is the variable approaching xx. Substituting f(x)=xf(x) = \sqrt{x} and f(z)=zf(z) = \sqrt{z} into the formula gives lim⁡z→xz−xz−x\lim_{z \to x} \frac{\sqrt{z} - \sqrt{x}}{z-x}.

Question 5

The value of lim⁡h→0tan⁡(h)h\lim_{h \to 0} \frac{\tan(h)}{h} can be found by recognizing it as the derivative of a certain function at a specific point. Which function and point is it?

  1. f(x)=sec⁡2(x)f(x) = \sec^2(x) at x=0x=0
  2. f(x)=tan⁡(x)f(x) = \tan(x) at x=0x=0 (correct answer)
  3. f(x)=tan⁡(x)f(x) = \tan(x) at x=π/4x=\pi/4
  4. f(x)=xf(x) = x at x=1x=1
Explanation: The limit lim⁡h→0tan⁡(h)h\lim_{h \to 0} \frac{\tan(h)}{h} can be rewritten as lim⁡h→0tan⁡(0+h)−tan⁡(0)h\lim_{h \to 0} \frac{\tan(0+h) - \tan(0)}{h} because tan⁡(0)=0\tan(0)=0. This expression matches the definition of f′(a)f'(a) for f(x)=tan⁡(x)f(x)=\tan(x) and a=0a=0. The value of this limit is f′(0)=sec⁡2(0)=1f'(0) = \sec^2(0) = 1.

Question 6

Let g(x)=x3g(x) = x^3. Which of the following limits represents g′(2)g'(2)?

  1. lim⁡h→0(2+h)3−8h\lim_{h \to 0} \frac{(2+h)^3 - 8}{h} (correct answer)
  2. lim⁡h→0h3−8h\lim_{h \to 0} \frac{h^3 - 8}{h}
  3. lim⁡x→2x3−2x−2\lim_{x \to 2} \frac{x^3 - 2}{x-2}
  4. lim⁡x→2x3−8x\lim_{x \to 2} \frac{x^3 - 8}{x}
Explanation: The derivative of a function g(x)g(x) at a point aa is defined as g′(a)=lim⁡h→0g(a+h)−g(a)hg'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h}. For g(x)=x3g(x)=x^3 and a=2a=2, we have g(2)=23=8g(2) = 2^3 = 8. Substituting into the definition gives g′(2)=lim⁡h→0g(2+h)−g(2)h=lim⁡h→0(2+h)3−8hg'(2) = \lim_{h \to 0} \frac{g(2+h) - g(2)}{h} = \lim_{h \to 0} \frac{(2+h)^3 - 8}{h}.

Question 7

If y=f(x)y = f(x), the notation dydx\frac{dy}{dx} is defined by which limit?

  1. lim⁡Δx→0f(x)−f(0)Δx\lim_{\Delta x \to 0} \frac{f(x) - f(0)}{\Delta x}
  2. lim⁡Δy→0ΔyΔx\lim_{\Delta y \to 0} \frac{\Delta y}{\Delta x}
  3. lim⁡x→0f(x)x\lim_{x \to 0} \frac{f(x)}{x}
  4. lim⁡Δx→0ΔyΔx\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} (correct answer)
Explanation: When you encounter questions about derivative notation, you're being tested on understanding the fundamental definition of a derivative as the limit of difference quotients. The derivative dydx\frac{dy}{dx} represents the instantaneous rate of change of yy with respect to xx. This is defined as the limit of the average rate of change as the change in xx approaches zero. If y=f(x)y = f(x), then when xx changes by Δx,y\Delta x,y changes by Δy=f(x+Δx)−f(x)\Delta y = f(x + \Delta x) - f(x). The average rate of change is ΔyΔx\frac{\Delta y}{\Delta x}, and taking the limit as Δx→0\Delta x \to 0 gives us the instantaneous rate of change: dydx=lim⁡Δx→0ΔyΔx\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}. This makes answer D correct. Answer A is incorrect because it compares f(x)f(x) to f(0)f(0) specifically, which would only give the average rate of change from x=0x = 0 to some point xx, not the instantaneous rate at a general point. Answer B has the limit variable wrong—we need Δx→0\Delta x \to 0, not Δy→0\Delta y \to 0. As Δx\Delta x shrinks, Δy\Delta y also approaches zero for continuous functions, but the limit must be taken with respect to the independent variable's change. Answer C represents lim⁡x→0f(x)x\lim_{x \to 0} \frac{f(x)}{x}, which could be related to the derivative at x=0x = 0 only if f(0)=0f(0) = 0, but this isn't the general definition. Remember: derivatives are always limits of difference quotients where the change in the independent variable approaches zero.

Question 8

Which of the following limits gives the slope of the tangent line to the graph of f(x)=ln⁡(x)f(x) = \ln(x) at the point (e,1)(e, 1)?

  1. lim⁡h→0ln⁡(e+h)−1h\lim_{h \to 0} \frac{\ln(e+h) - 1}{h} (correct answer)
  2. lim⁡h→0ln⁡(h)−1h\lim_{h \to 0} \frac{\ln(h) - 1}{h}
  3. lim⁡x→eln⁡(x)−ex−e\lim_{x \to e} \frac{\ln(x) - e}{x-e}
  4. lim⁡x→eln⁡(x)x−e\lim_{x \to e} \frac{\ln(x)}{x-e}
Explanation: The slope of the tangent line at a point (a,f(a))(a, f(a)) is given by the derivative f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. Here, the function is f(x)=ln⁡(x)f(x) = \ln(x) and the point is (e,1)(e, 1), so a=ea=e and f(a)=1f(a)=1. Substituting these values into the definition gives lim⁡h→0ln⁡(e+h)−1h\lim_{h \to 0} \frac{\ln(e+h) - 1}{h}.

Question 9

The expression lim⁡h→0cos⁡(π3+h)−12h\lim_{h \to 0} \frac{\cos(\frac{\pi}{3} + h) - \frac{1}{2}}{h} represents the derivative of a function f(x)f(x) at a point cc. What is the value of this limit?

  1. 12\frac{1}{2}
  2. 32\frac{\sqrt{3}}{2}
  3. −12-\frac{1}{2}
  4. −32-\frac{\sqrt{3}}{2} (correct answer)
Explanation: The given limit matches the definition of the derivative, f′(c)=lim⁡h→0f(c+h)−f(c)hf'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}. By comparing the expressions, we can identify the function f(x)=cos⁡(x)f(x) = \cos(x) and the point c=π3c = \frac{\pi}{3}, since f(π3)=cos⁡(π3)=12f(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}. The value of the limit is f′(π3)f'(\frac{\pi}{3}). The derivative of f(x)f(x) is f′(x)=−sin⁡(x)f'(x) = -\sin(x). Evaluating at c=π3c = \frac{\pi}{3} gives f′(π3)=−sin⁡(π3)=−32f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}.

Question 10

The value of the limit lim⁡x→axn−anx−a\lim_{x \to a} \frac{x^n - a^n}{x-a} represents the derivative of f(x)=xnf(x)=x^n at x=ax=a.

Using the principle described in the passage, what is the value of lim⁡x→1x100−1x2−1\lim_{x \to 1} \frac{x^{100} - 1}{x^2 - 1}?

  1. 100100
  2. 5050 (correct answer)
  3. 22
  4. 1/501/50
Explanation: The limit can be rewritten by factoring the denominator: lim⁡x→1x100−1x2−1=lim⁡x→1x100−1(x−1)(x+1)\lim_{x \to 1} \frac{x^{100} - 1}{x^2 - 1} = \lim_{x \to 1} \frac{x^{100} - 1}{(x - 1)(x + 1)}. We can split this into the product of two limits: (lim⁡x→1x100−1x−1)⋅(lim⁡x→11x+1)\left( \lim_{x \to 1} \frac{x^{100} - 1}{x - 1} \right) \cdot \left( \lim_{x \to 1} \frac{1}{x + 1} \right). The first limit, based on the provided principle, is the derivative of f(x)=x100f(x) = x^{100} at x=1x=1. The derivative is f′(x)=100x99f'(x) = 100x^{99}, so f′(1)=100(1)99=100f'(1) = 100(1)^{99} = 100. The second limit is 11+1=12\frac{1}{1+1} = \frac{1}{2}. Therefore, the value of the original limit is 100×12=50100 \times \frac{1}{2} = 50.

Question 11

The line tangent to the graph of a differentiable function f(x)f(x) at x=−1x=-1 is given by the equation y=4x+3y=4x+3. Which of the following statements must be true?

I. f(−1)=−1f(-1) = -1

II. lim⁡h→0f(−1+h)+1h=4\lim_{h \to 0} \frac{f(-1+h) + 1}{h} = 4

III. lim⁡x→−1f(x)+1x+1=4\lim_{x \to -1} \frac{f(x)+1}{x+1} = 4

  1. II only
  2. I and II only
  3. II and III only
  4. I, II, and III (correct answer)
Explanation: The equation of the tangent line to f(x)f(x) at x=ax=a is y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x-a). At x=−1x=-1, the tangent line must pass through the point (−1,f(−1))(-1, f(-1)). Plugging x=−1x=-1 into the line equation y=4x+3y=4x+3 gives y=4(−1)+3=−1y=4(-1)+3 = -1. So, f(−1)=−1f(-1)=-1. Statement I is true. The slope of the tangent line is the derivative at the point of tangency. The slope of y=4x+3y=4x+3 is 4, so f′(−1)=4f'(-1)=4. Statement II is lim⁡h→0f(−1+h)+1h\lim_{h \to 0} \frac{f(-1+h) + 1}{h}. Since f(−1)=−1f(-1)=-1, this is equivalent to lim⁡h→0f(−1+h)−f(−1)h\lim_{h \to 0} \frac{f(-1+h) - f(-1)}{h}, which is the definition of f′(−1)f'(-1). Since f′(−1)=4f'(-1)=4, statement II is true. Statement III is lim⁡x→−1f(x)+1x+1\lim_{x \to -1} \frac{f(x)+1}{x+1}. Since f(−1)=−1f(-1)=-1, this is lim⁡x→−1f(x)−f(−1)x−(−1)\lim_{x \to -1} \frac{f(x)-f(-1)}{x-(-1)}, which is the alternate definition of f′(−1)f'(-1). Since f′(−1)=4f'(-1)=4, statement III is true. Therefore, all three statements must be true.

Question 12

Consider the piecewise function f(x)={x2if x≤12x−1if x>1f(x) = \begin{cases} x^2 & \text{if } x \le 1 \\ 2x-1 & \text{if } x > 1 \end{cases}. To determine if f′(1)f'(1) exists, one must evaluate the left and right limits of f(1+h)−f(1)h\frac{f(1+h)-f(1)}{h} as h→0h \to 0. What can be concluded?

  1. The function continuity at x=1x=1 guarantees differentiability there.
  2. The left and right limits of the difference quotient are both equal to 2. (correct answer)
  3. The left limit of the difference quotient is 2, but the right limit is 1.
  4. The polynomial nature on both sides guarantees differentiability at x=1x=1.
Explanation: First, note f(1)=12=1f(1)=1^2=1. The left-hand derivative is lim⁡h→0−(1+h)2−1h=lim⁡h→0−1+2h+h2−1h=lim⁡h→0−(2+h)=2\lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{1+2h+h^2-1}{h} = \lim_{h \to 0^-} (2+h) = 2. The right-hand derivative is lim⁡h→0+(2(1+h)−1)−1h=lim⁡h→0+2+2h−2h=lim⁡h→0+2=2\lim_{h \to 0^+} \frac{(2(1+h)-1) - 1}{h} = \lim_{h \to 0^+} \frac{2+2h-2}{h} = \lim_{h \to 0^+} 2 = 2. Since the left-hand and right-hand limits of the difference quotient are equal, the derivative f′(1)f'(1) exists and equals 2.

Question 13

Let g(t)=e2tg(t) = e^{2t}. Which of the following expressions is equivalent to the instantaneous rate of change of gg at t=1t=1?

  1. ddtg(t)\frac{d}{dt}g(t)
  2. lim⁡h→0e2(t+h)−e2th\lim_{h \to 0} \frac{e^{2(t+h)} - e^{2t}}{h}
  3. e2(1+h)−e2h\frac{e^{2(1+h)} - e^2}{h}
  4. lim⁡t→1e2t−e2t−1\lim_{t \to 1} \frac{e^{2t} - e^2}{t-1} (correct answer)
Explanation: The instantaneous rate of change of a function g(t)g(t) at a specific point t=at=a is given by its derivative at that point, g′(a)g'(a). The definitions for g′(a)g'(a) are lim⁡h→0g(a+h)−g(a)h\lim_{h \to 0} \frac{g(a+h) - g(a)}{h} and lim⁡t→ag(t)−g(a)t−a\lim_{t \to a} \frac{g(t) - g(a)}{t-a}. For g(t)=e2tg(t)=e^{2t} at a=1a=1, g(1)=e2g(1)=e^2. The second definition becomes lim⁡t→1e2t−e2t−1\lim_{t \to 1} \frac{e^{2t} - e^2}{t-1}. Choice A is the derivative function, not its value at a point. Choice B is the limit definition of the derivative function g′(t)g'(t), not its value at t=1t=1. Choice C is the difference quotient for gg at t=1t=1, but without the limit, it only represents the average rate of change over an interval of length hh.

Question 14

Let f(x)=xcos⁡(x)f(x) = x \cos(x). Which of the following limits represents f′(π)f'(\pi)?

  1. lim⁡h→0(π+h)cos⁡(π+h)+πh\lim_{h \to 0} \frac{(\pi+h)\cos(\pi+h) + \pi}{h} (correct answer)
  2. lim⁡h→0πcos⁡(π+h)+πh\lim_{h \to 0} \frac{\pi\cos(\pi+h) + \pi}{h}
  3. lim⁡x→πxcos⁡(x)−πx−π\lim_{x \to \pi} \frac{x\cos(x) - \pi}{x-\pi}
  4. lim⁡h→0(x+h)cos⁡(x+h)−xcos⁡(x)h\lim_{h \to 0} \frac{(x+h)\cos(x+h) - x\cos(x)}{h}
Explanation: The definition of the derivative f′(a)f'(a) is lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. For the function f(x)=xcos⁡(x)f(x) = x\cos(x) at a=πa=\pi, we have f(π)=πcos⁡(π)=π(−1)=−πf(\pi) = \pi\cos(\pi) = \pi(-1) = -\pi. And f(π+h)=(π+h)cos⁡(π+h)f(\pi+h) = (\pi+h)\cos(\pi+h). Substituting these into the definition gives f′(π)=lim⁡h→0(π+h)cos⁡(π+h)−(−π)h=lim⁡h→0(π+h)cos⁡(π+h)+πhf'(\pi) = \lim_{h \to 0} \frac{(\pi+h)\cos(\pi+h) - (-\pi)}{h} = \lim_{h \to 0} \frac{(\pi+h)\cos(\pi+h) + \pi}{h}.

Question 15

The limit lim⁡h→0(1+2h)3−1h\lim_{h \to 0} \frac{(1+2h)^3 - 1}{h} represents the derivative of a function g(x)g(x) at a point x=ax=a. What is the value of this limit?

  1. 11
  2. 22
  3. 33
  4. 66 (correct answer)
Explanation: The limit lim⁡h→0(1+2h)3−1h\lim_{h \to 0} \frac{(1+2h)^3 - 1}{h} can be related to the derivative of f(x)=x3f(x)=x^3 at x=1x=1, which is f′(1)=lim⁡k→0(1+k)3−1kf'(1) = \lim_{k \to 0} \frac{(1+k)^3 - 1}{k}. Let k=2hk=2h. Then h=k/2h=k/2. As h→0h \to 0, k→0k \to 0. Substituting this into the given limit gives lim⁡k→0(1+k)3−1k/2=2lim⁡k→0(1+k)3−1k\lim_{k \to 0} \frac{(1+k)^3 - 1}{k/2} = 2 \lim_{k \to 0} \frac{(1+k)^3 - 1}{k}. This is 22 times the derivative of f(x)=x3f(x)=x^3 at x=1x=1. Since f′(x)=3x2f'(x)=3x^2, f′(1)=3(1)2=3f'(1)=3(1)^2=3. Therefore, the value of the limit is 2×3=62 \times 3 = 6. Alternatively, one can identify the limit as the derivative of g(x)=(1+2x)3g(x)=(1+2x)^3 at x=0x=0. By the chain rule, g′(x)=3(1+2x)2⋅2=6(1+2x)2g'(x) = 3(1+2x)^2 \cdot 2 = 6(1+2x)^2. Then g′(0)=6(1+0)2=6g'(0) = 6(1+0)^2 = 6.

Question 16

Let cc be a fixed real number and let the function gg be defined by g(x)=cos⁡(c)g(x) = \cos(c). Which expression represents g′(x)g'(x)?

  1. −sin⁡(c)-\sin(c)
  2. 00 (correct answer)
  3. lim⁡h→0cos⁡(c+h)−cos⁡(c)h\lim_{h \to 0} \frac{\cos(c+h) - \cos(c)}{h}
  4. lim⁡h→0cos⁡(x+h)−cos⁡(x)h\lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h}
Explanation: The function is g(x)=cos⁡(c)g(x) = \cos(c). Since cc is a fixed real number, cos⁡(c)\cos(c) is a constant value. The derivative of any constant function is zero. Therefore, g′(x)=0g'(x) = 0. By definition, g′(x)=lim⁡h→0g(x+h)−g(x)h=lim⁡h→0cos⁡(c)−cos⁡(c)h=lim⁡h→00h=0g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{\cos(c) - \cos(c)}{h} = \lim_{h \to 0} \frac{0}{h} = 0. Distractor A, −sin⁡(c)-\sin(c), would be the derivative of f(c)=cos⁡(c)f(c) = \cos(c) with respect to cc. Distractor C is a limit that evaluates to −sin⁡(c)-\sin(c), which is a constant, not the derivative function g′(x)g'(x). Distractor D defines −sin⁡(x)-\sin(x), which is the derivative of cos⁡(x)\cos(x), not cos⁡(c)\cos(c).

Question 17

The expression lim⁡h→0(3+h)2+7−4h\lim_{h \to 0} \frac{\sqrt{(3+h)^2 + 7} - 4}{h} represents the derivative of a function f(x)f(x) at a point x=ax=a. What are f(x)f(x) and aa?

  1. f(x)=x2+7f(x) = \sqrt{x^2+7} at a=3a=3 (correct answer)
  2. f(x)=x2f(x) = \sqrt{x^2} at a=3a=3
  3. f(x)=(3+x)2+7f(x) = \sqrt{(3+x)^2+7} at a=0a=0
  4. f(x)=x+7f(x) = \sqrt{x+7} at a=9a=9
Explanation: The limit definition of the derivative of a function f(x)f(x) at a point x=ax=a is f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. Comparing this to the given expression, we can identify a=3a=3. Then, f(a+h)=f(3+h)=(3+h)2+7f(a+h) = f(3+h) = \sqrt{(3+h)^2 + 7}. This implies that the function is f(x)=x2+7f(x) = \sqrt{x^2+7}. We must verify that the constant term matches f(a)=f(3)f(a) = f(3). Indeed, f(3)=32+7=9+7=16=4f(3) = \sqrt{3^2+7} = \sqrt{9+7} = \sqrt{16} = 4. Thus, the expression represents the derivative of f(x)=x2+7f(x) = \sqrt{x^2+7} at a=3a=3.

Question 18

What is the value of the limit lim⁡x→π/4tan⁡(x)−1x−π/4\lim_{x \to \pi/4} \frac{\tan(x) - 1}{x - \pi/4}?

  1. 11
  2. 22 (correct answer)
  3. 2\sqrt{2}
  4. π/4\pi/4
Explanation: The limit is in the form of the alternative definition of the derivative, f′(a)=lim⁡x→af(x)−f(a)x−af'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}. By comparing the given limit to this definition, we can identify f(x)=tan⁡(x)f(x) = \tan(x) and a=π/4a = \pi/4. We can verify that f(a)=f(π/4)=tan⁡(π/4)=1f(a) = f(\pi/4) = \tan(\pi/4) = 1, which matches the expression. The limit is therefore equal to the value of the derivative of f(x)f(x) at x=π/4x=\pi/4. The derivative of f(x)=tan⁡(x)f(x) = \tan(x) is f′(x)=sec⁡2(x)f'(x) = \sec^2(x). Evaluating this at x=π/4x=\pi/4 gives f′(π/4)=sec⁡2(π/4)=(2)2=2f'(\pi/4) = \sec^2(\pi/4) = (\sqrt{2})^2 = 2.

Question 19

If y=x3y = x^3, which of the following expressions is equivalent to dydx∣x=2\left. \frac{dy}{dx} \right|_{x=2}?

  1. (2.001)3−80.001\frac{(2.001)^3 - 8}{0.001}
  2. lim⁡h→0(2+h)3−8h\lim_{h \to 0} \frac{(2+h)^3 - 8}{h} (correct answer)
  3. lim⁡x→0(x+2)3−x32\lim_{x \to 0} \frac{(x+2)^3 - x^3}{2}
  4. lim⁡h→2(x+h)3−x3h\lim_{h \to 2} \frac{(x+h)^3 - x^3}{h}
Explanation: The notation dydx∣x=2\left. \frac{dy}{dx} \right|_{x=2} represents the derivative of the function y=f(x)=x3y=f(x)=x^3 evaluated at the point x=2x=2. This is denoted as f′(2)f'(2). The formal definition of the derivative at a point aa is f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. For f(x)=x3f(x)=x^3 and a=2a=2, this becomes f′(2)=lim⁡h→0(2+h)3−23h=lim⁡h→0(2+h)3−8hf'(2) = \lim_{h \to 0} \frac{(2+h)^3 - 2^3}{h} = \lim_{h \to 0} \frac{(2+h)^3 - 8}{h}. Choice A is a numerical approximation of the derivative, not the exact value defined by the limit. Choices C and D misuse the structure of the limit definition.

Question 20

If ff is a differentiable function, which of the following expressions is always equivalent to f′(c)f'(c)?

  1. lim⁡h→0f(c+h)−f(c−h)2h\lim_{h \to 0} \frac{f(c+h) - f(c-h)}{2h} (correct answer)
  2. lim⁡h→0f(c)−f(c+h)h\lim_{h \to 0} \frac{f(c) - f(c+h)}{h}
  3. lim⁡x→cf(x)−f(c)c−x\lim_{x \to c} \frac{f(x) - f(c)}{c-x}
  4. lim⁡h→0f(c+2h)−f(c)h\lim_{h \to 0} \frac{f(c+2h) - f(c)}{h}
Explanation: This expression is the symmetric difference quotient. We can rewrite it as lim⁡h→0f(c+h)−f(c)+f(c)−f(c−h)2h=lim⁡h→012(f(c+h)−f(c)h+f(c)−f(c−h)h)\lim_{h \to 0} \frac{f(c+h) - f(c) + f(c) - f(c-h)}{2h} = \lim_{h \to 0} \frac{1}{2} \left( \frac{f(c+h)-f(c)}{h} + \frac{f(c)-f(c-h)}{h} \right). The first part limits to f′(c)f'(c). For the second part, let k=−hk=-h. As h→0h \to 0, k→0k \to 0. It becomes lim⁡k→0f(c)−f(c+k)−k=lim⁡k→0f(c+k)−f(c)k=f′(c)\lim_{k \to 0} \frac{f(c)-f(c+k)}{-k} = \lim_{k \to 0} \frac{f(c+k)-f(c)}{k} = f'(c). So the limit is 12(f′(c)+f′(c))=f′(c)\frac{1}{2}(f'(c) + f'(c)) = f'(c). B is −f′(c)-f'(c). C is −f′(c)-f'(c). D is 2f′(c)2f'(c) by a change of variables k=2hk=2h.