Calculus 1 Quiz: Derivatives Of Sin And Cos
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Derivatives Of Sin And CosQuestion 1 of 20

For 0≤x≤2π0 \le x \le 2\pi, find the sum of all values of xx where the tangent lines to f(x)=sin⁡(x)f(x) = \sin(x) and g(x)=cos⁡(x)g(x) = \cos(x) are parallel.

3π/23\pi/2
5π/25\pi/2
2π2\pi
4π4\pi
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Calculus 1 Quiz

Calculus 1 Quiz: Derivatives Of Sin And Cos

Practice Derivatives Of Sin And Cos 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 Derivatives Of Sin And Cos, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

For 0≤x≤2π0 \le x \le 2\pi, find the sum of all values of xx where the tangent lines to f(x)=sin⁡(x)f(x) = \sin(x) and g(x)=cos⁡(x)g(x) = \cos(x) are parallel.

  1. 3π/23\pi/2
  2. 5π/25\pi/2 (correct answer)
  3. 2π2\pi
  4. 4π4\pi
Explanation: When you encounter a problem about parallel tangent lines, you're looking for points where two functions have equal derivatives, since parallel lines have identical slopes. To find where f(x)=sin⁡(x)f(x) = \sin(x) and g(x)=cos⁡(x)g(x) = \cos(x) have parallel tangent lines, you need to solve f′(x)=g′(x)f'(x) = g'(x). Taking derivatives: f′(x)=cos⁡(x)f'(x) = \cos(x) and g′(x)=−sin⁡(x)g'(x) = -\sin(x). Setting them equal: cos⁡(x)=−sin⁡(x)\cos(x) = -\sin(x). Dividing both sides by cos⁡(x)\cos(x) (being careful about when cos⁡(x)=0\cos(x) = 0): 1=−tan⁡(x)1 = -\tan(x), so tan⁡(x)=−1\tan(x) = -1. Over the interval [0,2π][0, 2\pi], this occurs at x=3π4x = \frac{3\pi}{4} and x=7π4x = \frac{7\pi}{4}. You can verify that cos⁡(x)≠0\cos(x) \neq 0 at these points, so our division was valid. The sum is 3π4+7π4=10π4=5π2\frac{3\pi}{4} + \frac{7\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}, which is choice B. Choice A (3π2\frac{3\pi}{2}) only accounts for one solution rather than both. Choice C (2π2\pi) might come from incorrectly thinking the solutions are at π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}, but tan⁡(x)\tan(x) is undefined at these points. Choice D (4π4\pi) could result from doubling the correct answer or finding incorrect solutions. Remember: when finding parallel tangent lines between two curves, always set their derivatives equal and solve completely over the given interval. Don't forget to check that your algebra steps are valid at the solutions you find.

Question 2

For which of the following values of xx does the graph of y=x+2cos⁡(x)y = x + 2\cos(x) have a horizontal tangent line?

  1. π/6\pi/6 (correct answer)
  2. π/3\pi/3
  3. π/2\pi/2
  4. 00
Explanation: A horizontal tangent line occurs where the derivative is equal to zero. First, find the derivative of yy with respect to xx. dydx=ddx(x+2cos⁡(x))=1−2sin⁡(x)\frac{dy}{dx} = \frac{d}{dx}(x + 2\cos(x)) = 1 - 2\sin(x) Set the derivative to zero and solve for xx: 1−2sin⁡(x)=01 - 2\sin(x) = 0 1=2sin⁡(x)1 = 2\sin(x) sin⁡(x)=1/2\sin(x) = 1/2 The principal value for xx in the first quadrant that satisfies this equation is x=π/6x = \pi/6. Other solutions exist, such as 5π/65\pi/6, but π/6\pi/6 is one of the choices.

Question 3

What is the 101st derivative of f(x)=cos⁡(x)f(x) = \cos(x)?

  1. sin⁡(x)\sin(x)
  2. −sin⁡(x)-\sin(x) (correct answer)
  3. cos⁡(x)\cos(x)
  4. −cos⁡(x)-\cos(x)
Explanation: The derivatives of cos⁡(x)\cos(x) follow a cycle of period 4: f′(x)=−sin⁡(x)f'(x) = -\sin(x) f′′(x)=−cos⁡(x)f''(x) = -\cos(x) f′′′(x)=sin⁡(x)f'''(x) = \sin(x) f(4)(x)=cos⁡(x)f^{(4)}(x) = \cos(x) To find the 101st derivative, we find the remainder of 101 divided by 4. 101=4×25+1101 = 4 \times 25 + 1. The remainder is 1. This means the 101st derivative is the same as the 1st derivative. Therefore, f(101)(x)=f′(x)=−sin⁡(x)f^{(101)}(x) = f'(x) = -\sin(x).

Question 4

A particle's position is described by p(t)=t−2sin⁡(t)p(t) = t - 2\sin(t). At what time tt in [0,π][0, \pi] is the particle momentarily at rest?

  1. 00
  2. π/6\pi/6
  3. π/2\pi/2
  4. π/3\pi/3 (correct answer)
Explanation: When you encounter a particle motion problem asking when the particle is "momentarily at rest," you need to find when the velocity equals zero. A particle is at rest when it's not moving, which means its instantaneous rate of change of position is zero. To find the velocity, you take the derivative of the position function: v(t)=p′(t)=ddt(t−2sin⁡(t))=1−2cos⁡(t)v(t) = p'(t) = \frac{d}{dt}(t - 2\sin(t)) = 1 - 2\cos(t) Setting the velocity equal to zero: 1−2cos⁡(t)=01 - 2\cos(t) = 0 Solving for tt: 2cos⁡(t)=12\cos(t) = 1, so cos⁡(t)=12\cos(t) = \frac{1}{2} In the interval [0,π][0, \pi], cosine equals 12\frac{1}{2} when t=π3t = \frac{\pi}{3}. This makes D the correct answer. Let's check why the other options are wrong. For A) t=0t = 0: cos⁡(0)=1\cos(0) = 1, so v(0)=1−2(1)=−1≠0v(0) = 1 - 2(1) = -1 \neq 0. For B) t=π6t = \frac{\pi}{6}: cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}, so v(π6)=1−2(32)=1−3≠0v(\frac{\pi}{6}) = 1 - 2(\frac{\sqrt{3}}{2}) = 1 - \sqrt{3} \neq 0. For C) t=π2t = \frac{\pi}{2}: cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0, so v(π2)=1−2(0)=1≠0v(\frac{\pi}{2}) = 1 - 2(0) = 1 \neq 0. Remember: "at rest" always means velocity equals zero, not position equals zero. Always differentiate the position function to find velocity, then set that derivative equal to zero and solve.

Question 5

Let f(x)=g(x)cos⁡(x)f(x) = g(x)\cos(x). Given that g(π/2)=2g(\pi/2) = 2 and g′(π/2)=−3g'(\pi/2) = -3, find f′(π/2)f'(\pi/2).

  1. -5
  2. -3
  3. 0
  4. -2 (correct answer)
Explanation: When you encounter a function that's the product of two functions, like f(x)=g(x)cos⁡(x)f(x) = g(x)\cos(x), you need the product rule to find its derivative. The product rule states that if f(x)=u(x)⋅v(x)f(x) = u(x) \cdot v(x), then f′(x)=u′(x)⋅v(x)+u(x)⋅v′(x)f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x). Here, u(x)=g(x)u(x) = g(x) and v(x)=cos⁡(x)v(x) = \cos(x), so f′(x)=g′(x)cos⁡(x)+g(x)⋅(−sin⁡(x))f'(x) = g'(x)\cos(x) + g(x) \cdot (-\sin(x)). Now you can substitute x=π/2x = \pi/2: f′(π/2)=g′(π/2)cos⁡(π/2)+g(π/2)⋅(−sin⁡(π/2))f'(\pi/2) = g'(\pi/2)\cos(\pi/2) + g(\pi/2) \cdot (-\sin(\pi/2)) Since cos⁡(π/2)=0\cos(\pi/2) = 0 and sin⁡(π/2)=1\sin(\pi/2) = 1: f′(π/2)=g′(π/2)⋅0+g(π/2)⋅(−1)=0+2⋅(−1)=−2f'(\pi/2) = g'(\pi/2) \cdot 0 + g(\pi/2) \cdot (-1) = 0 + 2 \cdot (-1) = -2 This confirms answer D is correct. Looking at the wrong answers: A) -5 might result from incorrectly adding g′(π/2)+g(π/2)=−3+2=−1g'(\pi/2) + g(\pi/2) = -3 + 2 = -1, then making an error with the trigonometric values. B) -3 comes from simply using g′(π/2)g'(\pi/2) without applying the product rule at all. C) 0 results from forgetting the second term of the product rule and only calculating g′(π/2)cos⁡(π/2)=−3⋅0=0g'(\pi/2)\cos(\pi/2) = -3 \cdot 0 = 0. Remember: when differentiating products, always write out the complete product rule formula before substituting values. The trigonometric values at special angles like π/2\pi/2 often simplify the calculation, but don't skip steps in your setup.

Question 6

Let h(x)=f(sin⁡(x))h(x) = f(\sin(x)). If f′(0)=4f'(0) = 4, what is h′(π)h'(\pi)?

  1. 0
  2. -4 (correct answer)
  3. 4
  4. 4cos⁡(1)4\cos(1)
Explanation: We must use the chain rule to find h′(x)h'(x). Let u=sin⁡(x)u = \sin(x). h′(x)=f′(sin⁡(x))⋅ddx(sin⁡(x))h'(x) = f'(\sin(x)) \cdot \frac{d}{dx}(\sin(x)) h′(x)=f′(sin⁡(x))⋅cos⁡(x)h'(x) = f'(\sin(x)) \cdot \cos(x) Now, we evaluate h′(x)h'(x) at x=πx = \pi. h′(π)=f′(sin⁡(π))⋅cos⁡(π)h'(\pi) = f'(\sin(\pi)) \cdot \cos(\pi) Since sin⁡(π)=0\sin(\pi) = 0 and cos⁡(π)=−1\cos(\pi) = -1: h′(π)=f′(0)⋅(−1)h'(\pi) = f'(0) \cdot (-1) We are given that f′(0)=4f'(0) = 4. Therefore: h′(π)=4⋅(−1)=−4h'(\pi) = 4 \cdot (-1) = -4.

Question 7

Evaluate the limit: lim⁡h→0cos⁡(π3+h)−12h\lim_{h \to 0} \frac{\cos(\frac{\pi}{3} + h) - \frac{1}{2}}{h}

  1. 32\frac{\sqrt{3}}{2}
  2. −32-\frac{\sqrt{3}}{2} (correct answer)
  3. 12\frac{1}{2}
  4. −12-\frac{1}{2}
Explanation: This limit represents the definition of the derivative of a function f(x)f(x) at a point x=ax=a, which is given by lim⁡h→0f(a+h)−f(a)h=f′(a)\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a). By comparing the given limit to this definition, we can identify f(x)=cos⁡(x)f(x) = \cos(x) and a=π3a = \frac{\pi}{3}. So, the value of the limit is equal to f′(π3)f'(\frac{\pi}{3}). First, find the derivative of f(x)f(x): f′(x)=−sin⁡(x)f'(x) = -\sin(x). Then, evaluate the derivative at x=π3x = \frac{\pi}{3}: f′(π3)=−sin⁡(π3)=−32f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}.

Question 8

The equation sin⁡(y)+cos⁡(x)=1\sin(y) + \cos(x) = 1 defines yy implicitly as a function of xx. Find dydx\frac{dy}{dx} at the point (π/2,0)(\pi/2, 0).

  1. 1 (correct answer)
  2. -1
  3. 0
  4. π/2\pi/2
Explanation: Differentiate both sides of the equation with respect to xx, remembering to use the chain rule for the term involving yy. ddx(sin⁡(y))+ddx(cos⁡(x))=ddx(1)\frac{d}{dx}(\sin(y)) + \frac{d}{dx}(\cos(x)) = \frac{d}{dx}(1) cos⁡(y)⋅dydx−sin⁡(x)=0\cos(y) \cdot \frac{dy}{dx} - \sin(x) = 0 Now, solve for dydx\frac{dy}{dx}: cos⁡(y)dydx=sin⁡(x)\cos(y) \frac{dy}{dx} = \sin(x) dydx=sin⁡(x)cos⁡(y)\frac{dy}{dx} = \frac{\sin(x)}{\cos(y)} Finally, substitute the coordinates of the point (π/2,0)(\pi/2, 0) into this expression: dydx∣(π/2,0)=sin⁡(π/2)cos⁡(0)=11=1\frac{dy}{dx} \bigg|_{(\pi/2, 0)} = \frac{\sin(\pi/2)}{\cos(0)} = \frac{1}{1} = 1.

Question 9

If g(x)=cos⁡(x)1−sin⁡(x)g(x) = \frac{\cos(x)}{1 - \sin(x)}, what is the value of g′(0)g'(0)?

  1. 1 (correct answer)
  2. -1
  3. 0
  4. 2
Explanation: To find g′(0)g'(0), we use the quotient rule: (uv)′=u′v−uv′v2(\frac{u}{v})' = \frac{u'v - uv'}{v^2}, with u=cos⁡(x)u = \cos(x) and v=1−sin⁡(x)v = 1 - \sin(x). u′=−sin⁡(x)u' = -\sin(x) and v′=−cos⁡(x)v' = -\cos(x). g′(x)=(−sin⁡(x))(1−sin⁡(x))−(cos⁡(x))(−cos⁡(x))(1−sin⁡(x))2g'(x) = \frac{(-\sin(x))(1 - \sin(x)) - (\cos(x))(-\cos(x))}{(1 - \sin(x))^2} g′(x)=−sin⁡(x)+sin⁡2(x)+cos⁡2(x)(1−sin⁡(x))2g'(x) = \frac{-\sin(x) + \sin^2(x) + \cos^2(x)}{(1 - \sin(x))^2} Using the identity sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1: g′(x)=1−sin⁡(x)(1−sin⁡(x))2=11−sin⁡(x)g'(x) = \frac{1 - \sin(x)}{(1 - \sin(x))^2} = \frac{1}{1 - \sin(x)} Now, evaluate at x=0x = 0: g′(0)=11−sin⁡(0)=11−0=1g'(0) = \frac{1}{1 - \sin(0)} = \frac{1}{1 - 0} = 1.

Question 10

What is the equation of the normal line to the graph of f(x)=cos⁡(x)f(x) = \cos(x) at x=π/2x = \pi/2?

  1. y=x−π/2y = x - \pi/2 (correct answer)
  2. y=−x+π/2y = -x + \pi/2
  3. y=1y = 1
  4. y=xy = x
Explanation: First, find the point of tangency. At x=π/2x = \pi/2, y=f(π/2)=cos⁡(π/2)=0y = f(\pi/2) = \cos(\pi/2) = 0. The point is (π/2,0)(\pi/2, 0). Next, find the slope of the tangent line by finding the derivative f′(x)f'(x). f′(x)=−sin⁡(x)f'(x) = -\sin(x) Evaluate the slope at x=π/2x = \pi/2: mtan=f′(π/2)=−sin⁡(π/2)=−1m_{tan} = f'(\pi/2) = -\sin(\pi/2) = -1. The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent's slope. mnormal=−1mtan=−1−1=1m_{normal} = -\frac{1}{m_{tan}} = -\frac{1}{-1} = 1. Finally, use the point-slope form to find the equation of the normal line: y−y1=m(x−x1)y - y_1 = m(x - x_1). y−0=1(x−π/2)y - 0 = 1(x - \pi/2) y=x−π/2y = x - \pi/2.

Question 11

Let f(x)=x2cos⁡(x)f(x) = x^2 \cos(x). What is the second derivative, f′′(x)f''(x)?

  1. (2−x2)cos⁡(x)−4xsin⁡(x)(2-x^2)\cos(x) - 4x\sin(x) (correct answer)
  2. (2+x2)cos⁡(x)−4xsin⁡(x)(2+x^2)\cos(x) - 4x\sin(x)
  3. 2cos⁡(x)−2xsin⁡(x)2\cos(x) - 2x\sin(x)
  4. 2xcos⁡(x)−x2sin⁡(x)2x\cos(x) - x^2\sin(x)
Explanation: To find the second derivative, we must apply the product rule twice. First, find the first derivative f′(x)f'(x): f′(x)=(ddxx2)cos⁡(x)+x2(ddxcos⁡(x))=2xcos⁡(x)−x2sin⁡(x)f'(x) = (\frac{d}{dx}x^2)\cos(x) + x^2(\frac{d}{dx}\cos(x)) = 2x\cos(x) - x^2\sin(x). Next, differentiate f′(x)f'(x) term by term, applying the product rule to each term: f′′(x)=[(ddx2x)cos⁡(x)+2x(ddxcos⁡(x))]−[(ddxx2)sin⁡(x)+x2(ddxsin⁡(x))]f''(x) = [(\frac{d}{dx}2x)\cos(x) + 2x(\frac{d}{dx}\cos(x))] - [(\frac{d}{dx}x^2)\sin(x) + x^2(\frac{d}{dx}\sin(x))] f′′(x)=[2cos⁡(x)+2x(−sin⁡(x))]−[2xsin⁡(x)+x2(cos⁡(x))]f''(x) = [2\cos(x) + 2x(-\sin(x))] - [2x\sin(x) + x^2(\cos(x))] f′′(x)=2cos⁡(x)−2xsin⁡(x)−2xsin⁡(x)−x2cos⁡(x)f''(x) = 2\cos(x) - 2x\sin(x) - 2x\sin(x) - x^2\cos(x) Combine like terms: f′′(x)=(2−x2)cos⁡(x)−4xsin⁡(x)f''(x) = (2-x^2)\cos(x) - 4x\sin(x).

Question 12

What is the equation of the line tangent to the graph of f(x)=cos⁡(2x)−sin⁡(x)f(x) = \cos(2x) - \sin(x) at x=π6x = \frac{\pi}{6}?

  1. y=−332(x−π6)y = -\frac{3\sqrt{3}}{2}(x - \frac{\pi}{6}) (correct answer)
  2. y=−3(x−π6)y = -\sqrt{3}(x - \frac{\pi}{6})
  3. y=332(x−π6)y = \frac{3\sqrt{3}}{2}(x - \frac{\pi}{6})
  4. y=0y = 0
Explanation: First, find the point of tangency. The y-coordinate is f(π6)=cos⁡(2⋅π6)−sin⁡(π6)=cos⁡(π3)−sin⁡(π6)=12−12=0f(\frac{\pi}{6}) = \cos(2\cdot\frac{\pi}{6}) - \sin(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) - \sin(\frac{\pi}{6}) = \frac{1}{2} - \frac{1}{2} = 0. The point is (π6,0)(\frac{\pi}{6}, 0). Next, find the slope of the tangent line by finding the derivative, f′(x)f'(x). Using the chain rule for cos⁡(2x)\cos(2x), we get: f′(x)=−sin⁡(2x)⋅2−cos⁡(x)=−2sin⁡(2x)−cos⁡(x)f'(x) = -\sin(2x) \cdot 2 - \cos(x) = -2\sin(2x) - \cos(x). Now, evaluate the slope at x=π6x = \frac{\pi}{6}: m=f′(π6)=−2sin⁡(2⋅π6)−cos⁡(π6)=−2sin⁡(π3)−cos⁡(π6)=−2(32)−32=−3−32=−332m = f'(\frac{\pi}{6}) = -2\sin(2\cdot\frac{\pi}{6}) - \cos(\frac{\pi}{6}) = -2\sin(\frac{\pi}{3}) - \cos(\frac{\pi}{6}) = -2(\frac{\sqrt{3}}{2}) - \frac{\sqrt{3}}{2} = -\sqrt{3} - \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2}. Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1), the equation is y−0=−332(x−π6)y - 0 = -\frac{3\sqrt{3}}{2}(x - \frac{\pi}{6}).

Question 13

Let g(x)=sin⁡(πcos⁡(x))g(x) = \sin(\pi \cos(x)). What is the value of g′(π2)g'(\frac{\pi}{2})?

  1. −π-\pi (correct answer)
  2. −1-1
  3. 11
  4. π\pi
Explanation: This requires the chain rule. Let the outer function be f(u)=sin⁡(u)f(u) = \sin(u) and the inner function be u(x)=πcos⁡(x)u(x) = \pi \cos(x). The derivative is g′(x)=f′(u(x))⋅u′(x)g'(x) = f'(u(x)) \cdot u'(x). f′(u)=cos⁡(u)f'(u) = \cos(u), so f′(u(x))=cos⁡(πcos⁡(x))f'(u(x)) = \cos(\pi \cos(x)). u′(x)=ddx(πcos⁡(x))=−πsin⁡(x)u'(x) = \frac{d}{dx}(\pi \cos(x)) = -\pi \sin(x). Combining these, we get: g′(x)=cos⁡(πcos⁡(x))⋅(−πsin⁡(x))g'(x) = \cos(\pi \cos(x)) \cdot (-\pi \sin(x)). Now, evaluate at x=π2x = \frac{\pi}{2}: g′(π2)=cos⁡(πcos⁡(π2))⋅(−πsin⁡(π2))g'(\frac{\pi}{2}) = \cos(\pi \cos(\frac{\pi}{2})) \cdot (-\pi \sin(\frac{\pi}{2})). Since cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0 and sin⁡(π2)=1\sin(\frac{\pi}{2}) = 1, we have: g′(π2)=cos⁡(π⋅0)⋅(−π⋅1)=cos⁡(0)⋅(−π)=1⋅(−π)=−πg'(\frac{\pi}{2}) = \cos(\pi \cdot 0) \cdot (-\pi \cdot 1) = \cos(0) \cdot (-\pi) = 1 \cdot (-\pi) = -\pi.

Question 14

The function f(x)=excos⁡(x)f(x) = e^x \cos(x) is decreasing on which of the following intervals?

  1. (0,π2)(0, \frac{\pi}{2})
  2. (−3π4,π4)-\frac{3\pi}{4}, \frac{\pi}{4})
  3. (π4,5π4)\frac{\pi}{4}, \frac{5\pi}{4}) (correct answer)
  4. (5π4,9π4)\frac{5\pi}{4}, \frac{9\pi}{4})
Explanation: A function is decreasing where its derivative is negative. First, find f′(x)f'(x) using the product rule: f′(x)=(ddxex)cos⁡(x)+ex(ddxcos⁡(x))=excos⁡(x)+ex(−sin⁡(x))=ex(cos⁡(x)−sin⁡(x))f'(x) = (\frac{d}{dx}e^x)\cos(x) + e^x(\frac{d}{dx}\cos(x)) = e^x \cos(x) + e^x(-\sin(x)) = e^x(\cos(x) - \sin(x)). To find where f(x)f(x) is decreasing, we need to solve f′(x)<0f'(x) < 0. Since exe^x is always positive, we need to solve cos⁡(x)−sin⁡(x)<0\cos(x) - \sin(x) < 0, which is equivalent to cos⁡(x)<sin⁡(x)\cos(x) < \sin(x). On the unit circle, sin⁡(x)=cos⁡(x)\sin(x) = \cos(x) at x=π4x = \frac{\pi}{4} and x=5π4x = \frac{5\pi}{4}. The inequality cos⁡(x)<sin⁡(x)\cos(x) < \sin(x) holds for xx values between these two points. Thus, the function is decreasing on the interval (π4,5π4)(\frac{\pi}{4}, \frac{5\pi}{4}).

Question 15

Let f(x)=sin⁡(x)1+cos⁡(x)f(x) = \frac{\sin(x)}{1+\cos(x)}. Find the value of f′(π3)f'(\frac{\pi}{3}).

  1. −23-\frac{2}{3}
  2. 00
  3. 33\frac{\sqrt{3}}{3}
  4. 23\frac{2}{3} (correct answer)
Explanation: We use the quotient rule: f′(x)=(ddxsin⁡x)(1+cos⁡x)−(sin⁡x)(ddx(1+cos⁡x))(1+cos⁡x)2f'(x) = \frac{(\frac{d}{dx}\sin x)(1+\cos x) - (\sin x)(\frac{d}{dx}(1+\cos x))}{(1+\cos x)^2}. f′(x)=(cos⁡x)(1+cos⁡x)−(sin⁡x)(−sin⁡x)(1+cos⁡x)2f'(x) = \frac{(\cos x)(1+\cos x) - (\sin x)(-\sin x)}{(1+\cos x)^2} f′(x)=cos⁡x+cos⁡2x+sin⁡2x(1+cos⁡x)2f'(x) = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2} Using the identity cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, we can simplify the numerator: f′(x)=cos⁡x+1(1+cos⁡x)2=11+cos⁡xf'(x) = \frac{\cos x + 1}{(1+\cos x)^2} = \frac{1}{1+\cos x}. Now, evaluate at x=π3x = \frac{\pi}{3}: f′(π3)=11+cos⁡(π3)=11+12=132=23f'(\frac{\pi}{3}) = \frac{1}{1+\cos(\frac{\pi}{3})} = \frac{1}{1+\frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}.

Question 16

A function is defined as f(x)=asin⁡(x)+bcos⁡(x)f(x) = a \sin(x) + b \cos(x) for constants aa and bb. If f′(π4)=2f'(\frac{\pi}{4}) = \sqrt{2} and f′′(π4)=−32f''(\frac{\pi}{4}) = -3\sqrt{2}, what is the value of f(π4)f(\frac{\pi}{4})?

  1. 2\sqrt{2}
  2. 222\sqrt{2}
  3. 323\sqrt{2} (correct answer)
  4. 424\sqrt{2}
Explanation: First, find the first and second derivatives of f(x)f(x): f′(x)=acos⁡(x)−bsin⁡(x)f'(x) = a \cos(x) - b \sin(x) f′′(x)=−asin⁡(x)−bcos⁡(x)f''(x) = -a \sin(x) - b \cos(x) Now, use the given information. At x=π4x = \frac{\pi}{4}, we know sin⁡(π4)=cos⁡(π4)=22\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}. f′(π4)=a(22)−b(22)=2f'(\frac{\pi}{4}) = a(\frac{\sqrt{2}}{2}) - b(\frac{\sqrt{2}}{2}) = \sqrt{2}. Dividing by 22\frac{\sqrt{2}}{2} gives a−b=2a - b = 2. f′′(π4)=−a(22)−b(22)=−32f''(\frac{\pi}{4}) = -a(\frac{\sqrt{2}}{2}) - b(\frac{\sqrt{2}}{2}) = -3\sqrt{2}. Dividing by 22\frac{\sqrt{2}}{2} gives −a−b=−6-a - b = -6, or a+b=6a + b = 6. We have a system of two linear equations:
  1. a−b=2a - b = 2
  2. a+b=6a + b = 6 Adding the two equations gives 2a=82a = 8, so a=4a = 4. Substituting into the second equation gives 4+b=64 + b = 6, so b=2b = 2. Now we need to find f(π4)f(\frac{\pi}{4}): f(π4)=asin⁡(π4)+bcos⁡(π4)=4(22)+2(22)=22+2=32f(\frac{\pi}{4}) = a \sin(\frac{\pi}{4}) + b \cos(\frac{\pi}{4}) = 4(\frac{\sqrt{2}}{2}) + 2(\frac{\sqrt{2}}{2}) = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2}.

Question 17

What is the slope of the tangent line to the curve y=cos⁡3(x)y = \cos^3(x) at x=π6x = \frac{\pi}{6}?

  1. 94\frac{9}{4}
  2. −18-\frac{1}{8}
  3. 98\frac{9}{8}
  4. −98-\frac{9}{8} (correct answer)
Explanation: When you see a question asking for the slope of a tangent line, you're being tested on derivatives and the chain rule. The slope of the tangent line at any point is simply the derivative of the function evaluated at that point. To find the derivative of y=cos⁡3(x)y = \cos^3(x), you need the chain rule since this is a composite function. Let u=cos⁡(x)u = \cos(x), so y=u3y = u^3. Using the chain rule: dydx=dydu⋅dudx=3u2⋅(−sin⁡(x))=3cos⁡2(x)⋅(−sin⁡(x))=−3cos⁡2(x)sin⁡(x)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \cdot (-\sin(x)) = 3\cos^2(x) \cdot (-\sin(x)) = -3\cos^2(x)\sin(x) Now evaluate this derivative at x=π6x = \frac{\pi}{6}:
  • cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
  • sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
So: dydx∣x=π6=−3(32)2⋅12=−3⋅34⋅12=−98\frac{dy}{dx}\bigg|_{x=\frac{\pi}{6}} = -3\left(\frac{\sqrt{3}}{2}\right)^2 \cdot \frac{1}{2} = -3 \cdot \frac{3}{4} \cdot \frac{1}{2} = -\frac{9}{8} This confirms answer D is correct. Looking at the wrong answers: A) 94\frac{9}{4} likely comes from forgetting the negative sign and miscalculating the fraction. B) −18-\frac{1}{8} suggests errors in both the chain rule application and trigonometric values. C) 98\frac{9}{8} gets the magnitude right but misses the crucial negative sign from the derivative of cosine. Study tip: Always remember that ddx[cos⁡(x)]=−sin⁡(x)\frac{d}{dx}[\cos(x)] = -\sin(x), and when using the chain rule, carefully track each step to avoid sign errors and computational mistakes.

Question 18

For which of the following values of xx in the interval [0,2π][0, 2\pi] does the function f(x)=x−2cos⁡(x)f(x) = x - 2\cos(x) have a horizontal tangent line?

  1. x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6}
  2. x=7π6x = \frac{7\pi}{6} and x=11π6x = \frac{11\pi}{6} (correct answer)
  3. x=4π3x = \frac{4\pi}{3} and x=5π3x = \frac{5\pi}{3}
  4. x=0x = 0 and x=πx = \pi
Explanation: A horizontal tangent line occurs where the derivative is equal to zero. First, find the derivative of f(x)f(x): f′(x)=ddx(x−2cos⁡(x))=1−2(−sin⁡(x))=1+2sin⁡(x)f'(x) = \frac{d}{dx}(x - 2\cos(x)) = 1 - 2(-\sin(x)) = 1 + 2\sin(x). Set the derivative equal to zero and solve for xx: 1+2sin⁡(x)=01 + 2\sin(x) = 0 2sin⁡(x)=−12\sin(x) = -1 sin⁡(x)=−12\sin(x) = -\frac{1}{2} In the interval [0,2π][0, 2\pi], the sine function is negative in the third and fourth quadrants. The reference angle is π6\frac{\pi}{6}. The solution in the third quadrant is x=π+π6=7π6x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}. The solution in the fourth quadrant is x=2π−π6=11π6x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}.

Question 19

If the equation ysin⁡(x)=cos⁡(y)y\sin(x) = \cos(y) defines yy as a differentiable function of xx, what is the value of dydx\frac{dy}{dx} at the point (0,π2)(0, \frac{\pi}{2})?

  1. −π2-\frac{\pi}{2} (correct answer)
  2. π2\frac{\pi}{2}
  3. −2π-\frac{2}{\pi}
  4. 2π\frac{2}{\pi}
Explanation: We use implicit differentiation with respect to xx. Apply the product rule to the left side and the chain rule to the right side: ddx(ysin⁡(x))=ddx(cos⁡(y))\frac{d}{dx}(y\sin(x)) = \frac{d}{dx}(\cos(y)) (dydx)sin⁡(x)+y(cos⁡(x))=−sin⁡(y)(dydx)(\frac{dy}{dx})\sin(x) + y(\cos(x)) = -\sin(y)(\frac{dy}{dx}) Now, we solve for dydx\frac{dy}{dx}: (dydx)sin⁡(x)+sin⁡(y)(dydx)=−ycos⁡(x)(\frac{dy}{dx})\sin(x) + \sin(y)(\frac{dy}{dx}) = -y\cos(x) dydx(sin⁡(x)+sin⁡(y))=−ycos⁡(x)\frac{dy}{dx}(\sin(x) + \sin(y)) = -y\cos(x) dydx=−ycos⁡(x)sin⁡(x)+sin⁡(y)\frac{dy}{dx} = \frac{-y\cos(x)}{\sin(x) + \sin(y)} Substitute the point (x,y)=(0,π2)(x, y) = (0, \frac{\pi}{2}) into the expression: dydx=−(π2)cos⁡(0)sin⁡(0)+sin⁡(π2)=−(π2)(1)0+1=−π2\frac{dy}{dx} = \frac{-(\frac{\pi}{2})\cos(0)}{\sin(0) + \sin(\frac{\pi}{2})} = \frac{-(\frac{\pi}{2})(1)}{0 + 1} = -\frac{\pi}{2}.

Question 20

The position of a particle moving along the x-axis is given by s(t)=3sin⁡(t)−cos⁡(2t)s(t) = 3\sin(t) - \cos(2t) for t≥0t \ge 0. What is the acceleration of the particle at t=π2t = \frac{\pi}{2}?

  1. −7-7 (correct answer)
  2. −4-4
  3. 00
  4. 77
Explanation: Acceleration is the second derivative of the position function. First, we find the velocity function, v(t)=s′(t)v(t) = s'(t): v(t)=ddt(3sin⁡(t)−cos⁡(2t))=3cos⁡(t)−(−sin⁡(2t)⋅2)=3cos⁡(t)+2sin⁡(2t)v(t) = \frac{d}{dt}(3\sin(t) - \cos(2t)) = 3\cos(t) - (-\sin(2t) \cdot 2) = 3\cos(t) + 2\sin(2t). Next, we find the acceleration function, a(t)=v′(t)a(t) = v'(t): a(t)=ddt(3cos⁡(t)+2sin⁡(2t))=−3sin⁡(t)+2(cos⁡(2t)⋅2)=−3sin⁡(t)+4cos⁡(2t)a(t) = \frac{d}{dt}(3\cos(t) + 2\sin(2t)) = -3\sin(t) + 2(\cos(2t) \cdot 2) = -3\sin(t) + 4\cos(2t). Finally, we evaluate the acceleration at t=π2t = \frac{\pi}{2}: a(π2)=−3sin⁡(π2)+4cos⁡(2⋅π2)=−3sin⁡(π2)+4cos⁡(π)=−3(1)+4(−1)=−3−4=−7a(\frac{\pi}{2}) = -3\sin(\frac{\pi}{2}) + 4\cos(2 \cdot \frac{\pi}{2}) = -3\sin(\frac{\pi}{2}) + 4\cos(\pi) = -3(1) + 4(-1) = -3 - 4 = -7.