Calculus 1 Quiz: Derivatives Of Tan And Other Trig
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Derivatives Of Tan And Other TrigQuestion 1 of 20

Find the x-coordinates in the interval (0,π)(0, \pi) where the tangent line to f(x)=2tan⁡(x)f(x) = 2\tan(x) is parallel to the line y=4x−1y=4x-1.

x=π/3x = \pi/3 only
x=π/4x = \pi/4 and x=3π/4x = 3\pi/4
x=π/3x = \pi/3 and x=2π/3x = 2\pi/3
x=π/6x = \pi/6 and x=5π/6x = 5\pi/6
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Calculus 1 Quiz

Calculus 1 Quiz: Derivatives Of Tan And Other Trig

Practice Derivatives Of Tan And Other Trig in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Derivatives Of Tan And Other Trig, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Find the x-coordinates in the interval (0,π)(0, \pi) where the tangent line to f(x)=2tan⁡(x)f(x) = 2\tan(x) is parallel to the line y=4x−1y=4x-1.

  1. x=π/3x = \pi/3 only
  2. x=π/4x = \pi/4 and x=3π/4x = 3\pi/4 (correct answer)
  3. x=π/3x = \pi/3 and x=2π/3x = 2\pi/3
  4. x=π/6x = \pi/6 and x=5π/6x = 5\pi/6
Explanation: The tangent line is parallel to y=4x−1y=4x-1 if its slope is 4. The slope is given by the derivative f′(x)f'(x). The derivative of f(x)=2tan⁡(x)f(x)=2\tan(x) is f′(x)=2sec⁡2(x)f'(x) = 2\sec^2(x). Set the derivative equal to 4: 2sec⁡2(x)=42\sec^2(x) = 4, which simplifies to sec⁡2(x)=2\sec^2(x) = 2. Taking the square root gives sec⁡(x)=±2\sec(x) = \pm\sqrt{2}, which is equivalent to cos⁡(x)=±12\cos(x) = \pm\frac{1}{\sqrt{2}}. In the interval (0,π)(0, \pi), the solutions are x=π4x = \frac{\pi}{4} (where cos⁡(x)=12\cos(x) = \frac{1}{\sqrt{2}}) and x=3π4x = \frac{3\pi}{4} (where cos⁡(x)=−12\cos(x) = -\frac{1}{\sqrt{2}}).

Question 2

Let g(x)=cot⁡(x)g(x) = \cot(x). Find the second derivative, g′′(x)g''(x).

  1. 2csc⁡(x)cot⁡(x)2\csc(x)\cot(x)
  2. −2csc⁡2(x)cot⁡(x)-2\csc^2(x)\cot(x)
  3. 2csc⁡2(x)cot⁡(x)2\csc^2(x)\cot(x) (correct answer)
  4. −csc⁡2(x)+cot⁡2(x)-\csc^2(x) + \cot^2(x)
Explanation: First, find the first derivative: g′(x)=−csc⁡2(x)g'(x) = -\csc^2(x). To find the second derivative, we differentiate g′(x)g'(x). This requires the chain rule. Let u=csc⁡(x)u = \csc(x), so g′(x)=−u2g'(x) = -u^2. Then g′′(x)=−2u⋅u′g''(x) = -2u \cdot u'. The derivative of u=csc⁡(x)u = \csc(x) is u′=−csc⁡(x)cot⁡(x)u' = -\csc(x)\cot(x). Substituting back, we get g′′(x)=−2(csc⁡(x))(−csc⁡(x)cot⁡(x))=2csc⁡2(x)cot⁡(x)g''(x) = -2(\csc(x))(-\csc(x)\cot(x)) = 2\csc^2(x)\cot(x).

Question 3

If f(x)=cot⁡(e2x)f(x) = \cot(e^{2x}), what is f′(x)f'(x)?

  1. −2e2xcsc⁡2(e2x)-2e^{2x} \csc^2(e^{2x}) (correct answer)
  2. 2e2xcsc⁡2(e2x)2e^{2x} \csc^2(e^{2x})
  3. −e2xcsc⁡2(e2x)-e^{2x} \csc^2(e^{2x})
  4. 2e2xsec⁡2(e2x)2e^{2x} \sec^2(e^{2x})
Explanation: This requires the chain rule. Let u=e2xu = e^{2x}. Then f(u)=cot⁡(u)f(u) = \cot(u). The derivative of cot⁡(u)\cot(u) with respect to uu is −csc⁡2(u)-\csc^2(u). The derivative of u=e2xu = e^{2x} with respect to xx is dudx=e2x⋅2=2e2x\frac{du}{dx} = e^{2x} \cdot 2 = 2e^{2x}. By the chain rule, f′(x)=dfdu⋅dudx=−csc⁡2(e2x)⋅2e2x=−2e2xcsc⁡2(e2x)f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = -\csc^2(e^{2x}) \cdot 2e^{2x} = -2e^{2x} \csc^2(e^{2x}).

Question 4

For which value of xx in the interval (−π/2,π/2)(-\pi/2, \pi/2) is the tangent line to the graph of f(x)=tan⁡(x)−4xf(x) = \tan(x) - 4x horizontal?

  1. π6\frac{\pi}{6}
  2. π4\frac{\pi}{4}
  3. π3\frac{\pi}{3} (correct answer)
  4. arccos⁡(14)\arccos(\frac{1}{4})
Explanation: A horizontal tangent line occurs when the derivative is zero. First, find the derivative: f′(x)=sec⁡2(x)−4f'(x) = \sec^2(x) - 4. Set the derivative to zero and solve for xx: sec⁡2(x)−4=0\sec^2(x) - 4 = 0, which means sec⁡2(x)=4\sec^2(x) = 4. Taking the square root gives sec⁡(x)=±2\sec(x) = \pm 2, which is equivalent to cos⁡(x)=±12\cos(x) = \pm \frac{1}{2}. In the interval (−π/2,π/2)(-\pi/2, \pi/2), cos⁡(x)\cos(x) is positive. So we need to solve cos⁡(x)=12\cos(x) = \frac{1}{2}. The solution in this interval is x=π3x = \frac{\pi}{3}.

Question 5

If sec⁡(xy)=y\sec(xy) = y, find dydx\frac{dy}{dx} at the point (π,−1)(\pi, -1).

  1. 00 (correct answer)
  2. 1/π1/\pi
  3. −1/π-1/\pi
  4. π\pi
Explanation: Differentiate both sides implicitly with respect to xx. The left side requires the chain rule and product rule: sec⁡(xy)tan⁡(xy)⋅(1⋅y+xdydx)=dydx\sec(xy)\tan(xy) \cdot (1\cdot y + x \frac{dy}{dx}) = \frac{dy}{dx}. Now, substitute the point x=π,y=−1x=\pi, y=-1. The argument of the trig functions is xy=−πxy = -\pi. sec⁡(−π)=−1\sec(-\pi) = -1 and tan⁡(−π)=0\tan(-\pi) = 0. Substituting these values into the differentiated equation gives: (−1)(0)⋅(−1+πdydx)=dydx(-1)(0) \cdot (-1 + \pi \frac{dy}{dx}) = \frac{dy}{dx}. This simplifies to 0=dydx0 = \frac{dy}{dx}.

Question 6

If f(x)=sec⁡(x)cot⁡(x)f(x) = \sec(x)\cot(x), find f′(x)f'(x).

  1. sec⁡2(x)\sec^2(x)
  2. sec⁡(x)tan⁡(x)\sec(x)\tan(x)
  3. −csc⁡(x)cot⁡(x)-\csc(x)\cot(x) (correct answer)
  4. csc⁡2(x)\csc^2(x)
Explanation: The easiest approach is to first simplify the function f(x)f(x) using trigonometric identities. f(x)=sec⁡(x)cot⁡(x)=1cos⁡(x)⋅cos⁡(x)sin⁡(x)=1sin⁡(x)=csc⁡(x)f(x) = \sec(x)\cot(x) = \frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)} = \csc(x). Now, differentiate the simplified function: f′(x)=(csc⁡(x))′=−csc⁡(x)cot⁡(x)f'(x) = (\csc(x))' = -\csc(x)\cot(x). Alternatively, using the product rule on the original function yields the same result after simplification.

Question 7

What is the derivative of f(x)=tan⁡(2x)x2+1f(x) = \frac{\tan(2x)}{x^2+1}?

  1. 2sec⁡2(2x)(x2+1)−2xtan⁡(2x)(x2+1)2\frac{2\sec^2(2x)(x^2+1) - 2x\tan(2x)}{(x^2+1)^2} (correct answer)
  2. sec⁡2(2x)(x2+1)−2xtan⁡(2x)(x2+1)2\frac{\sec^2(2x)(x^2+1) - 2x\tan(2x)}{(x^2+1)^2}
  3. 2sec⁡2(2x)2x\frac{2\sec^2(2x)}{2x}
  4. 2xtan⁡(2x)−2sec⁡2(2x)(x2+1)(x2+1)2\frac{2x\tan(2x) - 2\sec^2(2x)(x^2+1)}{(x^2+1)^2}
Explanation: Use the quotient rule (uv)′=u′v−uv′v2(\frac{u}{v})' = \frac{u'v - uv'}{v^2} with u=tan⁡(2x)u = \tan(2x) and v=x2+1v = x^2+1. First, find the derivatives: u′=sec⁡2(2x)⋅2=2sec⁡2(2x)u' = \sec^2(2x) \cdot 2 = 2\sec^2(2x) (by the chain rule) and v′=2xv' = 2x. Now, substitute into the quotient rule formula: f′(x)=(2sec⁡2(2x))(x2+1)−(tan⁡(2x))(2x)(x2+1)2=2sec⁡2(2x)(x2+1)−2xtan⁡(2x)(x2+1)2f'(x) = \frac{(2\sec^2(2x))(x^2+1) - (\tan(2x))(2x)}{(x^2+1)^2} = \frac{2\sec^2(2x)(x^2+1) - 2x\tan(2x)}{(x^2+1)^2}.

Question 8

What is the smallest positive value of xx for which the function f(x)=tan⁡(x)−4xf(x) = \tan(x) - 4x has a horizontal tangent line?

  1. 2π/32\pi/3
  2. π/6\pi/6
  3. π/3\pi/3 (correct answer)
  4. π/4\pi/4
Explanation: When you see a question asking for where a function has a horizontal tangent line, you're looking for points where the derivative equals zero. A horizontal tangent means the slope is zero at that point. To find where f(x)=tan⁡(x)−4xf(x) = \tan(x) - 4x has a horizontal tangent, you need to find f′(x)f'(x) and set it equal to zero. Taking the derivative: f′(x)=sec⁡2(x)−4f'(x) = \sec^2(x) - 4 Setting this equal to zero: sec⁡2(x)−4=0\sec^2(x) - 4 = 0, which gives us sec⁡2(x)=4\sec^2(x) = 4. Since sec⁡(x)=1cos⁡(x)\sec(x) = \frac{1}{\cos(x)}, we have 1cos⁡2(x)=4\frac{1}{\cos^2(x)} = 4, so cos⁡2(x)=14\cos^2(x) = \frac{1}{4}. This means cos⁡(x)=±12\cos(x) = \pm\frac{1}{2}. For positive values of x,cos⁡(x)=12x,\cos(x) = \frac{1}{2} at x=π3,5π3,...x = \frac{\pi}{3}, \frac{5\pi}{3}, ... and cos⁡(x)=−12\cos(x) = -\frac{1}{2} at x=2π3,4π3,...x = \frac{2\pi}{3}, \frac{4\pi}{3}, ... The smallest positive value is π3\frac{\pi}{3}, making C correct. Looking at the wrong answers: A) 2π3\frac{2\pi}{3} is where cos⁡(x)=−12\cos(x) = -\frac{1}{2}, which does satisfy our equation, but it's larger than π3\frac{\pi}{3}. B) At π6,cos⁡(x)=32\frac{\pi}{6},\cos(x) = \frac{\sqrt{3}}{2}, so sec⁡2(x)=43≠4\sec^2(x) = \frac{4}{3} \neq 4. D) At π4,cos⁡(x)=22\frac{\pi}{4},\cos(x) = \frac{\sqrt{2}}{2}, so sec⁡2(x)=2≠4\sec^2(x) = 2 \neq 4. Remember: horizontal tangents occur where the derivative equals zero. Always check your trigonometric values carefully and find the smallest positive solution when asked.

Question 9

If y=csc⁡(x)x2y = \frac{\csc(x)}{x^2}, then dydx=\frac{dy}{dx} =?

  1. −csc⁡(x)(xcot⁡(x)+2)x3\frac{-\csc(x)(x\cot(x) + 2)}{x^3} (correct answer)
  2. csc⁡(x)(xcot⁡(x)−2)x3\frac{\csc(x)(x\cot(x) - 2)}{x^3}
  3. −csc⁡(x)cot⁡(x)2x\frac{-\csc(x)\cot(x)}{2x}
  4. csc⁡(x)(2−xcot⁡(x))x3\frac{\csc(x)(2 - x\cot(x))}{x^3}
Explanation: Using the quotient rule, (fg)′=f′g−fg′g2(\frac{f}{g})' = \frac{f'g - fg'}{g^2}, with f(x)=csc⁡(x)f(x) = \csc(x) and g(x)=x2g(x) = x^2. We have f′(x)=−csc⁡(x)cot⁡(x)f'(x) = -\csc(x)\cot(x) and g′(x)=2xg'(x) = 2x. Plugging these into the formula gives dydx=(−csc⁡(x)cot⁡(x))(x2)−(csc⁡(x))(2x)(x2)2=−x2csc⁡(x)cot⁡(x)−2xcsc⁡(x)x4\frac{dy}{dx} = \frac{(-\csc(x)\cot(x))(x^2) - (\csc(x))(2x)}{(x^2)^2} = \frac{-x^2\csc(x)\cot(x) - 2x\csc(x)}{x^4}. Factoring −xcsc⁡(x)-x\csc(x) from the numerator gives −xcsc⁡(x)(xcot⁡(x)+2)x4\frac{-x\csc(x)(x\cot(x) + 2)}{x^4}. Simplifying by canceling an xx gives −csc⁡(x)(xcot⁡(x)+2)x3\frac{-\csc(x)(x\cot(x) + 2)}{x^3}.

Question 10

Find the equation of the line normal to the graph of f(x)=sec⁡(x)f(x) = \sec(x) at the point where x=π/3x = \pi/3.

  1. y−2=23(x−π3)y - 2 = 2\sqrt{3}(x - \frac{\pi}{3})
  2. y−2=−123(x−π3)y - 2 = -\frac{1}{2\sqrt{3}}(x - \frac{\pi}{3}) (correct answer)
  3. y−2=−23(x−π3)y - 2 = -2\sqrt{3}(x - \frac{\pi}{3})
  4. y−12=−123(x−π3)y - \frac{1}{2} = -\frac{1}{2\sqrt{3}}(x - \frac{\pi}{3})
Explanation: The normal line is perpendicular to the tangent line. First, find the slope of the tangent line, mtanm_{tan}, by evaluating the derivative at x=π/3x = \pi/3. f′(x)=sec⁡(x)tan⁡(x)f'(x) = \sec(x)\tan(x). At x=π/3x=\pi/3, mtan=sec⁡(π/3)tan⁡(π/3)=(2)(3)=23m_{tan} = \sec(\pi/3)\tan(\pi/3) = (2)(\sqrt{3}) = 2\sqrt{3}. The slope of the normal line, mnormm_{norm}, is the negative reciprocal of the tangent slope: mnorm=−1mtan=−123m_{norm} = -\frac{1}{m_{tan}} = -\frac{1}{2\sqrt{3}}. The point of interest is (π/3,f(π/3))=(π/3,sec⁡(π/3))=(π/3,2)(\pi/3, f(\pi/3)) = (\pi/3, \sec(\pi/3)) = (\pi/3, 2). Using the point-slope form, the equation of the normal line is y−2=−123(x−π3)y - 2 = -\frac{1}{2\sqrt{3}}(x - \frac{\pi}{3}).

Question 11

The angle θ\theta of a swinging pendulum is described by θ(t)=0.2cot⁡(πt)\theta(t) = 0.2 \cot(\pi t), where tt is time in seconds. What is the instantaneous rate of change of the angle with respect to time at t=14t = \frac{1}{4} seconds?

  1. −0.4-0.4
  2. 0.4π0.4\pi
  3. −0.8π-0.8\pi
  4. −0.4π-0.4\pi (correct answer)
Explanation: The instantaneous rate of change is the derivative, θ′(t)\theta'(t). Using the chain rule, θ′(t)=0.2⋅(−csc⁡2(πt))⋅(π)=−0.2πcsc⁡2(πt)\theta'(t) = 0.2 \cdot (-\csc^2(\pi t)) \cdot (\pi) = -0.2\pi \csc^2(\pi t). Now, evaluate at t=14t = \frac{1}{4}. We have πt=π4\pi t = \frac{\pi}{4}. So, θ′(14)=−0.2πcsc⁡2(π4)=−0.2π(2)2=−0.2π(2)=−0.4π\theta'(\frac{1}{4}) = -0.2\pi \csc^2(\frac{\pi}{4}) = -0.2\pi (\sqrt{2})^2 = -0.2\pi(2) = -0.4\pi.

Question 12

Let f(x)=ktan⁡(x)f(x) = k \tan(x), where kk is a constant. If the slope of the tangent line to f(x)f(x) at x=π4x = \frac{\pi}{4} is 8, what is the value of kk?

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 16
Explanation: The slope of the tangent line is given by the derivative f′(x)f'(x). First, find the derivative: f′(x)=ksec⁡2(x)f'(x) = k \sec^2(x). The problem states that the slope at x=π4x = \frac{\pi}{4} is 8. So, f′(π4)=8f'(\frac{\pi}{4}) = 8. We have ksec⁡2(π4)=8k \sec^2(\frac{\pi}{4}) = 8. Since sec⁡(π4)=2\sec(\frac{\pi}{4}) = \sqrt{2}, we have sec⁡2(π4)=2\sec^2(\frac{\pi}{4}) = 2. The equation becomes k(2)=8k(2) = 8, which implies k=4k=4.

Question 13

Evaluate the limit: lim⁡x→π/2tan⁡(2x)cot⁡(x)\lim_{x \to \pi/2} \frac{\tan(2x)}{\cot(x)}

  1. 22
  2. −2-2 (correct answer)
  3. 11
  4. Does not exist
Explanation: When you encounter limits involving trigonometric functions that create indeterminate forms, the key strategy is to rewrite the expression using fundamental trigonometric identities before applying limit properties. Let's rewrite this expression using basic trig identities. Since tan⁡(2x)=sin⁡(2x)cos⁡(2x)\tan(2x) = \frac{\sin(2x)}{\cos(2x)} and cot⁡(x)=cos⁡(x)sin⁡(x)\cot(x) = \frac{\cos(x)}{\sin(x)}, we have: tan⁡(2x)cot⁡(x)=sin⁡(2x)cos⁡(2x)⋅sin⁡(x)cos⁡(x)=sin⁡(2x)sin⁡(x)cos⁡(2x)cos⁡(x)\frac{\tan(2x)}{\cot(x)} = \frac{\sin(2x)}{\cos(2x)} \cdot \frac{\sin(x)}{\cos(x)} = \frac{\sin(2x)\sin(x)}{\cos(2x)\cos(x)} Using the double angle identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x): sin⁡(2x)sin⁡(x)cos⁡(2x)cos⁡(x)=2sin⁡(x)cos⁡(x)sin⁡(x)cos⁡(2x)cos⁡(x)=2sin⁡2(x)cos⁡(2x)\frac{\sin(2x)\sin(x)}{\cos(2x)\cos(x)} = \frac{2\sin(x)\cos(x)\sin(x)}{\cos(2x)\cos(x)} = \frac{2\sin^2(x)}{\cos(2x)} Since cos⁡(2x)=cos⁡2(x)−sin⁡2(x)=1−2sin⁡2(x)\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2\sin^2(x), we get: 2sin⁡2(x)1−2sin⁡2(x)\frac{2\sin^2(x)}{1 - 2\sin^2(x)} As x→π/2x \to \pi/2, we have sin⁡(x)→1\sin(x) \to 1, so sin⁡2(x)→1\sin^2(x) \to 1. Therefore: lim⁡x→π/22sin⁡2(x)1−2sin⁡2(x)=2(1)1−2(1)=2−1=−2\lim_{x \to \pi/2} \frac{2\sin^2(x)}{1 - 2\sin^2(x)} = \frac{2(1)}{1 - 2(1)} = \frac{2}{-1} = -2 Choice A (2) ignores the negative sign in the denominator. Choice C (1) likely comes from incorrectly simplifying the trigonometric identities. Choice D (does not exist) assumes the limit is undefined, but our algebraic manipulation shows it exists and equals -2. Study tip: When limits involve multiple trig functions, always convert to sines and cosines first, then apply double angle formulas to simplify before substituting the limit value.

Question 14

Let f(x)=ln⁡(sec⁡(x)+tan⁡(x))f(x) = \ln(\sec(x) + \tan(x)). What is f′(x)f'(x)?

  1. tan⁡(x)\tan(x)
  2. sec⁡(x)\sec(x) (correct answer)
  3. 1sec⁡(x)+tan⁡(x)\frac{1}{\sec(x) + \tan(x)}
  4. sec⁡(x)tan⁡(x)+sec⁡2(x)\sec(x)\tan(x) + \sec^2(x)
Explanation: Using the chain rule, f′(x)=1sec⁡(x)+tan⁡(x)⋅ddx(sec⁡(x)+tan⁡(x))f'(x) = \frac{1}{\sec(x) + \tan(x)} \cdot \frac{d}{dx}(\sec(x) + \tan(x)). The derivative of the inner part is sec⁡(x)tan⁡(x)+sec⁡2(x)\sec(x)\tan(x) + \sec^2(x). So, f′(x)=sec⁡(x)tan⁡(x)+sec⁡2(x)sec⁡(x)+tan⁡(x)f'(x) = \frac{\sec(x)\tan(x) + \sec^2(x)}{\sec(x) + \tan(x)}. Factoring sec⁡(x)\sec(x) from the numerator gives f′(x)=sec⁡(x)(tan⁡(x)+sec⁡(x))sec⁡(x)+tan⁡(x)f'(x) = \frac{\sec(x)(\tan(x) + \sec(x))}{\sec(x) + \tan(x)}. The term (sec⁡(x)+tan⁡(x))(\sec(x) + \tan(x)) cancels, leaving f′(x)=sec⁡(x)f'(x) = \sec(x).

Question 15

The function f(x)=tan⁡(x)−xf(x) = \tan(x) - x is increasing on the interval (−π/2,π/2)(-\pi/2, \pi/2) when:

  1. f(x)>0f(x) > 0
  2. f′(x)>0f'(x) > 0 (correct answer)
  3. f′′(x)>0f''(x) > 0
  4. The function is always decreasing on this interval
Explanation: A function is increasing when its first derivative is positive. First, find the derivative: f′(x)=sec⁡2(x)−1f'(x) = \sec^2(x) - 1. Using the identity sec⁡2(x)−1=tan⁡2(x)\sec^2(x) - 1 = \tan^2(x), we have f′(x)=tan⁡2(x)f'(x) = \tan^2(x). Since the square of any real number is non-negative, tan⁡2(x)≥0\tan^2(x) \ge 0 for all xx in the domain. The function is increasing wherever f′(x)>0f'(x) > 0. This is true for all xx in (−π/2,π/2)(-\pi/2, \pi/2) except at x=0x=0, where f′(0)=0f'(0)=0. So the condition for increasing is f′(x)>0f'(x) > 0.

Question 16

If y=(1+tan⁡(x))3y = (1 + \tan(x))^3, find the slope of the tangent line at x=0x=0.

  1. 0
  2. 1
  3. 3 (correct answer)
  4. 9
Explanation: The slope of the tangent line is the value of the derivative at that point. We use the chain rule. Let u=1+tan⁡(x)u = 1 + \tan(x), so y=u3y = u^3. Then dydu=3u2\frac{dy}{du} = 3u^2 and dudx=sec⁡2(x)\frac{du}{dx} = \sec^2(x). So, dydx=3u2⋅sec⁡2(x)=3(1+tan⁡(x))2sec⁡2(x)\frac{dy}{dx} = 3u^2 \cdot \sec^2(x) = 3(1 + \tan(x))^2 \sec^2(x). Now, evaluate at x=0x=0. We have tan⁡(0)=0\tan(0) = 0 and sec⁡(0)=1\sec(0) = 1. Plugging these values in: slope=3(1+0)2(1)2=3(1)(1)=3\text{slope} = 3(1 + 0)^2 (1)^2 = 3(1)(1) = 3.

Question 17

Let f(x)=sec⁡(x)f(x) = \sec(x). 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 derivative: f′(x)=sec⁡(x)tan⁡(x)f'(x) = \sec(x)\tan(x). To find the second derivative, use the product rule: f′′(x)=(sec⁡(x)tan⁡(x))′=(sec⁡(x))′tan⁡(x)+sec⁡(x)(tan⁡(x))′=(sec⁡(x)tan⁡(x))tan⁡(x)+sec⁡(x)(sec⁡2(x))=sec⁡(x)tan⁡2(x)+sec⁡3(x)f''(x) = (\sec(x)\tan(x))' = (\sec(x))'\tan(x) + \sec(x)(\tan(x))' = (\sec(x)\tan(x))\tan(x) + \sec(x)(\sec^2(x)) = \sec(x)\tan^2(x) + \sec^3(x). Now, evaluate at x=π4x = \frac{\pi}{4}. At this point, tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1 and sec⁡(π4)=2\sec(\frac{\pi}{4}) = \sqrt{2}. So, f′′(π4)=(2)(1)2+(2)3=2+22=32f''(\frac{\pi}{4}) = (\sqrt{2})(1)^2 + (\sqrt{2})^3 = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}.

Question 18

Let f(x)=e−xcsc⁡(x)f(x) = e^{-x}\csc(x). Find f′(π/2)f'(\pi/2).

  1. −e−π/2-e^{-\pi/2} (correct answer)
  2. 00
  3. e−π/2e^{-\pi/2}
  4. −2e−π/2-2e^{-\pi/2}
Explanation: Use the product rule: f′(x)=(e−x)′csc⁡(x)+e−x(csc⁡(x))′=−e−xcsc⁡(x)+e−x(−csc⁡(x)cot⁡(x))=−e−xcsc⁡(x)(1+cot⁡(x))f'(x) = (e^{-x})'\csc(x) + e^{-x}(\csc(x))' = -e^{-x}\csc(x) + e^{-x}(-\csc(x)\cot(x)) = -e^{-x}\csc(x)(1 + \cot(x)). Now evaluate at x=π/2x = \pi/2. We have csc⁡(π/2)=1\csc(\pi/2) = 1 and cot⁡(π/2)=0\cot(\pi/2) = 0. Substituting these values: f′(π/2)=−e−π/2(1)(1+0)=−e−π/2f'(\pi/2) = -e^{-\pi/2}(1)(1+0) = -e^{-\pi/2}.

Question 19

Find dydx\frac{dy}{dx} for the curve defined by the equation tan⁡(y)=x2+sin⁡(x)\tan(y) = x^2 + \sin(x).

  1. 2x−cos⁡(x)sec⁡2(y)\frac{2x - \cos(x)}{\sec^2(y)}
  2. (2x+cos⁡(x))sec⁡2(y)(2x + \cos(x)) \sec^2(y)
  3. (2x+cos⁡(x))cos⁡2(y)(2x + \cos(x)) \cos^2(y) (correct answer)
  4. 2x+cos⁡(x)tan⁡(y)\frac{2x + \cos(x)}{\tan(y)}
Explanation: Use implicit differentiation. Differentiate both sides with respect to xx. The derivative of tan⁡(y)\tan(y) is sec⁡2(y)dydx\sec^2(y) \frac{dy}{dx} by the chain rule. The derivative of x2+sin⁡(x)x^2 + \sin(x) is 2x+cos⁡(x)2x + \cos(x). So, sec⁡2(y)dydx=2x+cos⁡(x)\sec^2(y) \frac{dy}{dx} = 2x + \cos(x). To solve for dydx\frac{dy}{dx}, divide by sec⁡2(y)\sec^2(y): dydx=2x+cos⁡(x)sec⁡2(y)\frac{dy}{dx} = \frac{2x + \cos(x)}{\sec^2(y)}. Since 1sec⁡2(y)=cos⁡2(y)\frac{1}{\sec^2(y)} = \cos^2(y), this can be rewritten as dydx=(2x+cos⁡(x))cos⁡2(y)\frac{dy}{dx} = (2x + \cos(x))\cos^2(y).

Question 20

Let h(x)=tan⁡(x2)h(x) = \tan(x^2). What is the slope of the tangent line to the graph of h(x)h(x) at x=π2x = \frac{\sqrt{\pi}}{2}?

  1. 2π2\sqrt{\pi} (correct answer)
  2. π\sqrt{\pi}
  3. 22
  4. 4π4\sqrt{\pi}
Explanation: The slope of the tangent line is given by the derivative h′(x)h'(x). Using the chain rule, let u=x2u=x^2, so dudx=2x\frac{du}{dx}=2x. Then h(u)=tan⁡(u)h(u)=\tan(u), so dhdu=sec⁡2(u)\frac{dh}{du}=\sec^2(u). Thus, h′(x)=sec⁡2(x2)⋅2xh'(x) = \sec^2(x^2) \cdot 2x. We need to evaluate this at x=π2x = \frac{\sqrt{\pi}}{2}. First, x2=(π2)2=π4x^2 = (\frac{\sqrt{\pi}}{2})^2 = \frac{\pi}{4}. Now substitute into the derivative: h′(π2)=sec⁡2(π4)⋅2(π2)=(2)2⋅π=2πh'(\frac{\sqrt{\pi}}{2}) = \sec^2(\frac{\pi}{4}) \cdot 2(\frac{\sqrt{\pi}}{2}) = (\sqrt{2})^2 \cdot \sqrt{\pi} = 2\sqrt{\pi}.