Calculus 1 Quiz: Derivatives Of Cos Sin Ex Lnx
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Derivatives Of Cos Sin Ex LnxQuestion 1 of 20

The tangent line to the graph of f(x)=sin⁡(x)+kexf(x) = \sin(x) + k e^x at x=0x=0 is parallel to the line y=3x−1y=3x-1. What is the value of kk?

11
33
22
44
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Calculus 1 Quiz: Derivatives Of Cos Sin Ex Lnx

Practice Derivatives Of Cos Sin Ex Lnx 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 Cos Sin Ex Lnx, 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

The tangent line to the graph of f(x)=sin⁡(x)+kexf(x) = \sin(x) + k e^x at x=0x=0 is parallel to the line y=3x−1y=3x-1. What is the value of kk?

  1. 11
  2. 33
  3. 22 (correct answer)
  4. 44
Explanation: When you see a question about tangent lines being parallel to other lines, you're working with derivatives and the geometric meaning of slope. Parallel lines have identical slopes, so you need to find the derivative of the given function and set it equal to the slope of the reference line. The line y=3x−1y = 3x - 1 has slope 3, so the tangent line to f(x)=sin⁡(x)+kexf(x) = \sin(x) + ke^x at x=0x = 0 must also have slope 3. To find this slope, take the derivative: f′(x)=cos⁡(x)+kexf'(x) = \cos(x) + ke^x. At x=0x = 0, this becomes f′(0)=cos⁡(0)+ke0=1+k⋅1=1+kf'(0) = \cos(0) + ke^0 = 1 + k \cdot 1 = 1 + k. Since the tangent line must have slope 3, you set 1+k=31 + k = 3, which gives k=2k = 2. Looking at the wrong answers: Choice A gives k=1k = 1, which would make the slope 1+1=21 + 1 = 2 instead of 3. Choice B gives k=3k = 3, resulting in slope 1+3=41 + 3 = 4, which is too steep. Choice D gives k=4k = 4, making the slope 1+4=51 + 4 = 5, even steeper than needed. Each of these represents the common error of forgetting that f′(0)=1+kf'(0) = 1 + k, not just kk. The correct answer is C) 2. Study tip: When finding tangent line slopes, always evaluate the derivative at the specific point given. Don't forget that cos⁡(0)=1\cos(0) = 1 and e0=1e^0 = 1 - these constants matter in your final calculation.

Question 2

Let h(x)=ecos⁡(2x)h(x) = e^{\cos(2x)}. What is the value of h′(π/4)h'(\pi/4)?

  1. -2 (correct answer)
  2. -1
  3. 1
  4. 2
Explanation: This requires the chain rule. The derivative of eue^u is eu⋅u′e^u \cdot u'. Here, u=cos⁡(2x)u = \cos(2x), and u′=−sin⁡(2x)⋅2u' = -\sin(2x) \cdot 2. So, h′(x)=ecos⁡(2x)⋅(−sin⁡(2x)⋅2)=−2sin⁡(2x)ecos⁡(2x)h'(x) = e^{\cos(2x)} \cdot (-\sin(2x) \cdot 2) = -2\sin(2x)e^{\cos(2x)}. Now, evaluate at x=π/4x = \pi/4: h′(π/4)=−2sin⁡(π/2)ecos⁡(π/2)=−2(1)e0=−2(1)(1)=−2h'(\pi/4) = -2\sin(\pi/2)e^{\cos(\pi/2)} = -2(1)e^0 = -2(1)(1) = -2.

Question 3

What is the smallest positive value of xx for which the derivatives of f(x)=sin⁡(2x)f(x) = \sin(2x) and g(x)=2sin⁡(x)g(x) = 2\sin(x) are equal?

  1. π/3\pi/3
  2. 2π/32\pi/3 (correct answer)
  3. π\pi
  4. 2π2\pi
Explanation: First, find the derivatives: f′(x)=2cos⁡(2x)f'(x) = 2\cos(2x) and g′(x)=2cos⁡(x)g'(x) = 2\cos(x). Set them equal: 2cos⁡(2x)=2cos⁡(x)2\cos(2x) = 2\cos(x), which simplifies to cos⁡(2x)=cos⁡(x)\cos(2x) = \cos(x). Using the double-angle identity cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1, we get 2cos⁡2(x)−1=cos⁡(x)2\cos^2(x) - 1 = \cos(x). This is a quadratic equation in cos⁡(x)\cos(x): 2cos⁡2(x)−cos⁡(x)−1=02\cos^2(x) - \cos(x) - 1 = 0. Factoring gives (2cos⁡(x)+1)(cos⁡(x)−1)=0(2\cos(x) + 1)(\cos(x) - 1) = 0. The solutions are cos⁡(x)=1\cos(x) = 1 (which gives x=2kπx=2k\pi) and cos⁡(x)=−1/2\cos(x) = -1/2 (which gives x=2π/3+2kπx = 2\pi/3 + 2k\pi and x=4π/3+2kπx = 4\pi/3 + 2k\pi). The smallest positive solution is x=2π/3x = 2\pi/3.

Question 4

For the function g(t)=et(sin⁡t+cos⁡t)g(t) = e^t (\sin t + \cos t), find a value of tt in the interval (0,2π)(0, 2\pi) for which the tangent line to the graph of gg is horizontal.

  1. π/4\pi/4
  2. 3π/43\pi/4
  3. π/2\pi/2 (correct answer)
  4. π\pi
Explanation: A horizontal tangent line occurs when the derivative is zero. Using the product rule: g′(t)=(et)(sin⁡t+cos⁡t)+et(cos⁡t−sin⁡t)g'(t) = (e^t)(\sin t + \cos t) + e^t(\cos t - \sin t). Factoring out ete^t gives g′(t)=et(sin⁡t+cos⁡t+cos⁡t−sin⁡t)=2etcos⁡tg'(t) = e^t(\sin t + \cos t + \cos t - \sin t) = 2e^t \cos t. Since ete^t is never zero, we must have cos⁡t=0\cos t = 0. In the interval (0,2π)(0, 2\pi), this occurs at t=π/2t = \pi/2 and t=3π/2t = 3\pi/2. Of the choices provided, π/2\pi/2 is a correct answer.

Question 5

Let h(x)=sin⁡xexh(x) = \frac{\sin x}{e^x}. What is the xx-coordinate of the first positive critical point of h(x)h(x)?

  1. π/4\pi/4 (correct answer)
  2. π/2\pi/2
  3. 3π/43\pi/4
  4. π\pi
Explanation: Critical points occur where the derivative is zero or undefined. We find the derivative using the quotient rule: h′(x)=(cos⁡x)(ex)−(sin⁡x)(ex)(ex)2=ex(cos⁡x−sin⁡x)e2x=cos⁡x−sin⁡xexh'(x) = \frac{(\cos x)(e^x) - (\sin x)(e^x)}{(e^x)^2} = \frac{e^x(\cos x - \sin x)}{e^{2x}} = \frac{\cos x - \sin x}{e^x}. To find critical points, we set h′(x)=0h'(x) = 0. Since exe^x is never zero, we must have cos⁡x−sin⁡x=0\cos x - \sin x = 0, which means cos⁡x=sin⁡x\cos x = \sin x. This is equivalent to tan⁡x=1\tan x = 1. The first positive value of xx for which this is true is x=π/4x = \pi/4.

Question 6

What is the equation of the tangent line to the graph of f(x)=exx+1f(x) = \frac{e^x}{x+1} at x=0x=0?

  1. y=x+1y = x + 1
  2. y=xy = x
  3. y=−x+1y = -x + 1
  4. y=1y = 1 (correct answer)
Explanation: When finding a tangent line equation, you need two key pieces: a point on the curve and the slope at that point. The tangent line formula is y−y1=m(x−x1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is your point and mm is the slope (derivative value). First, find the point on the curve at x=0x = 0: f(0)=e00+1=11=1f(0) = \frac{e^0}{0+1} = \frac{1}{1} = 1 So your point is (0,1)(0, 1). Next, find the slope by computing f′(0)f'(0). Using the quotient rule on f(x)=exx+1f(x) = \frac{e^x}{x+1}: f′(x)=(x+1)⋅ex−ex⋅1(x+1)2=ex(x+1)−ex(x+1)2=xex(x+1)2f'(x) = \frac{(x+1) \cdot e^x - e^x \cdot 1}{(x+1)^2} = \frac{e^x(x+1) - e^x}{(x+1)^2} = \frac{xe^x}{(x+1)^2} At x=0x = 0: f′(0)=0⋅e0(0+1)2=01=0f'(0) = \frac{0 \cdot e^0}{(0+1)^2} = \frac{0}{1} = 0 The tangent line equation becomes: y−1=0(x−0)y - 1 = 0(x - 0), which simplifies to y=1y = 1. This is answer D. Answer A (y=x+1y = x + 1) assumes the slope is 1 instead of 0. Answer B (y=xy = x) gets the slope wrong and passes through the wrong point. Answer C (y=−x+1y = -x + 1) incorrectly assumes a slope of -1, possibly from sign errors in the derivative calculation. Study tip: When the derivative equals zero at a point, the tangent line is horizontal. Always double-check your quotient rule algebra, as derivative errors are the most common mistake in tangent line problems.

Question 7

Let f(x)=sin⁡(x)+cos⁡(x)f(x) = \sin(x) + \cos(x). What is the 2023rd derivative, f(2023)(x)f^{(2023)}(x)?

  1. sin⁡(x)−cos⁡(x)\sin(x) - \cos(x) (correct answer)
  2. cos⁡(x)−sin⁡(x)\cos(x) - \sin(x)
  3. −sin⁡(x)−cos⁡(x)-\sin(x) - \cos(x)
  4. sin⁡(x)+cos⁡(x)\sin(x) + \cos(x)
Explanation: The derivatives of sin⁡(x)\sin(x) and cos⁡(x)\cos(x) follow a cycle of period 4. To find the 2023rd derivative, we need to find the remainder of 2023 divided by 4. 2023=4×505+32023 = 4 \times 505 + 3. So, the 2023rd derivative is the same as the 3rd derivative. f′(x)=cos⁡(x)−sin⁡(x)f'(x) = \cos(x) - \sin(x), f′′(x)=−sin⁡(x)−cos⁡(x)f''(x) = -\sin(x) - \cos(x), f′′′(x)=−cos⁡(x)+sin⁡(x)f'''(x) = -\cos(x) + \sin(x). Therefore, f(2023)(x)=sin⁡(x)−cos⁡(x)f^{(2023)}(x) = \sin(x) - \cos(x).

Question 8

Let f(x)=exf(x) = e^x and g(x)=x2+1g(x) = x^2 + 1. If h(x)=f(ln⁡(g(x)))h(x) = f(\ln(g(x))), what is the value of h′(1)h'(1)?

  1. ee
  2. e2e^2
  3. 22 (correct answer)
  4. ln⁡(2)\ln(2)
Explanation: The function h(x)h(x) can be simplified before differentiation. Since exe^x and ln⁡x\ln x are inverse functions, h(x)=f(ln⁡(g(x)))=eln⁡(g(x))=g(x)h(x) = f(\ln(g(x))) = e^{\ln(g(x))} = g(x). So, h(x)=x2+1h(x) = x^2 + 1. The derivative is h′(x)=2xh'(x) = 2x. Evaluating at x=1x=1 gives h′(1)=2(1)=2h'(1) = 2(1) = 2.

Question 9

If f(x)=sin⁡(ln⁡x)f(x) = \sin(\ln x), what is f′(e)f'(e)?

  1. 1e\frac{1}{e}
  2. cos⁡(1)e\frac{\cos(1)}{e} (correct answer)
  3. cos⁡(1)\cos(1)
  4. 00
Explanation: Using the chain rule, f′(x)=cos⁡(ln⁡x)⋅ddx(ln⁡x)=cos⁡(ln⁡x)⋅1xf'(x) = \cos(\ln x) \cdot \frac{d}{dx}(\ln x) = \cos(\ln x) \cdot \frac{1}{x}. Now, substitute x=ex = e: f′(e)=cos⁡(ln⁡e)⋅1e=cos⁡(1)⋅1e=cos⁡(1)ef'(e) = \cos(\ln e) \cdot \frac{1}{e} = \cos(1) \cdot \frac{1}{e} = \frac{\cos(1)}{e}.

Question 10

Let f(x)=Aex−Bcos⁡(x)f(x) = A e^x - B \cos(x). Given that f′(0)=3f'(0) = 3 and f′′(0)=5f''(0) = 5, what is the value of A+BA + B?

  1. 2
  2. 3
  3. 5 (correct answer)
  4. 8
Explanation: First, find the first and second derivatives of f(x)f(x). f′(x)=Aex+Bsin⁡(x)f'(x) = A e^x + B \sin(x). f′′(x)=Aex+Bcos⁡(x)f''(x) = A e^x + B \cos(x). Now use the given conditions. From f′(0)=3f'(0) = 3, we have Ae0+Bsin⁡(0)=3⇒A(1)+B(0)=3⇒A=3A e^0 + B \sin(0) = 3 \Rightarrow A(1) + B(0) = 3 \Rightarrow A = 3. From f′′(0)=5f''(0) = 5, we have Ae0+Bcos⁡(0)=5⇒A(1)+B(1)=5⇒A+B=5A e^0 + B \cos(0) = 5 \Rightarrow A(1) + B(1) = 5 \Rightarrow A + B = 5. The question asks for A+BA+B, which is directly given by the second condition.

Question 11

If y=cos⁡3(ln⁡x)y = \cos^3(\ln x), what is dydx\frac{dy}{dx}?

  1. −3xcos⁡2(ln⁡x)sin⁡(ln⁡x)-\frac{3}{x} \cos^2(\ln x) \sin(\ln x) (correct answer)
  2. 3xcos⁡2(ln⁡x)sin⁡(ln⁡x)\frac{3}{x} \cos^2(\ln x) \sin(\ln x)
  3. −3cos⁡2(ln⁡x)sin⁡(ln⁡x)-3 \cos^2(\ln x) \sin(\ln x)
  4. −3sin⁡(ln⁡x)x-\frac{3\sin(\ln x)}{x}
Explanation: This requires applying the chain rule multiple times. Let y=u3y = u^3, u=cos⁡(v)u = \cos(v), and v=ln⁡xv = \ln x. Then dydu=3u2\frac{dy}{du} = 3u^2, dudv=−sin⁡(v)\frac{du}{dv} = -\sin(v), and dvdx=1x\frac{dv}{dx} = \frac{1}{x}. By the chain rule, dydx=dydududvdvdx=3u2⋅(−sin⁡v)⋅1x\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx} = 3u^2 \cdot (-\sin v) \cdot \frac{1}{x}. Substituting back gives dydx=3cos⁡2(ln⁡x)⋅(−sin⁡(ln⁡x))⋅1x=−3xcos⁡2(ln⁡x)sin⁡(ln⁡x)\frac{dy}{dx} = 3\cos^2(\ln x) \cdot (-\sin(\ln x)) \cdot \frac{1}{x} = -\frac{3}{x} \cos^2(\ln x) \sin(\ln x).

Question 12

Let f(x)=xexf(x) = x e^x. What is the value of f′′(2)f''(2), the second derivative of ff evaluated at x=2x=2?

  1. 2e22e^2
  2. 3e23e^2
  3. 4e24e^2 (correct answer)
  4. e2e^2
Explanation: To find the second derivative, we must apply the product rule twice. The first derivative is f′(x)=(1)ex+x(ex)=ex(1+x)f'(x) = (1)e^x + x(e^x) = e^x(1+x). The second derivative is f′′(x)=(ex)(1+x)+(ex)(1)=ex(1+x+1)=ex(x+2)f''(x) = (e^x)(1+x) + (e^x)(1) = e^x(1+x+1) = e^x(x+2). Evaluating at x=2x=2, we get f′′(2)=e2(2+2)=4e2f''(2) = e^2(2+2) = 4e^2.

Question 13

At which value of xx does the graph of the function f(x)=ln⁡xxf(x) = \frac{\ln x}{x} have a horizontal tangent line?

  1. x=1x = 1
  2. x=ex = e (correct answer)
  3. x=1/ex = 1/e
  4. x=e2x = e^2
Explanation: A horizontal tangent line occurs where the derivative f′(x)f'(x) is equal to zero. Using the quotient rule, f′(x)=(1x)⋅x−(ln⁡x)⋅1x2=1−ln⁡xx2f'(x) = \frac{(\frac{1}{x}) \cdot x - (\ln x) \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}. Setting f′(x)=0f'(x) = 0 requires the numerator to be zero: 1−ln⁡x=01 - \ln x = 0, which implies ln⁡x=1\ln x = 1. The solution is x=ex = e.

Question 14

If ln⁡(y)+sin⁡(x)=x\ln(y) + \sin(x) = x, what is the value of dydx\frac{dy}{dx} at the point where x=π/2x = \pi/2?

  1. 00
  2. 11
  3. eπ/2−1e^{\pi/2 - 1} (correct answer)
  4. e1−π/2e^{1 - \pi/2}
Explanation: First, find the y-coordinate when x=π/2x=\pi/2: ln⁡(y)+sin⁡(π/2)=π/2⇒ln⁡(y)+1=π/2⇒ln⁡(y)=π/2−1⇒y=eπ/2−1\ln(y) + \sin(\pi/2) = \pi/2 \Rightarrow \ln(y) + 1 = \pi/2 \Rightarrow \ln(y) = \pi/2 - 1 \Rightarrow y = e^{\pi/2 - 1}. Next, use implicit differentiation on the original equation: 1ydydx+cos⁡(x)=1\frac{1}{y} \frac{dy}{dx} + \cos(x) = 1. Solve for dydx\frac{dy}{dx}: dydx=y(1−cos⁡(x))\frac{dy}{dx} = y(1 - \cos(x)). Substitute the point (x,y)=(π/2,eπ/2−1)(x,y) = (\pi/2, e^{\pi/2 - 1}): dydx=eπ/2−1(1−cos⁡(π/2))=eπ/2−1(1−0)=eπ/2−1\frac{dy}{dx} = e^{\pi/2 - 1}(1 - \cos(\pi/2)) = e^{\pi/2 - 1}(1 - 0) = e^{\pi/2 - 1}.

Question 15

Let g(x)=sin⁡(x)−cos⁡(x)g(x) = \sin(x) - \cos(x). Find g(2023)(x)g^{(2023)}(x), the 2023rd derivative of g(x)g(x).

  1. cos⁡(x)+sin⁡(x)\cos(x) + \sin(x)
  2. −sin⁡(x)+cos⁡(x)-\sin(x) + \cos(x)
  3. sin⁡(x)−cos⁡(x)\sin(x) - \cos(x)
  4. −cos⁡(x)−sin⁡(x)-\cos(x) - \sin(x) (correct answer)
Explanation: We find the first few derivatives of g(x)g(x) to identify a pattern: g′(x)=cos⁡(x)−(−sin⁡(x))=cos⁡(x)+sin⁡(x)g'(x) = \cos(x) - (-\sin(x)) = \cos(x) + \sin(x) g′′(x)=−sin⁡(x)+cos⁡(x)g''(x) = -\sin(x) + \cos(x) g′′′(x)=−cos⁡(x)−sin⁡(x)g'''(x) = -\cos(x) - \sin(x) g(4)(x)=sin⁡(x)−cos⁡(x)=g(x)g^{(4)}(x) = \sin(x) - \cos(x) = g(x) The derivatives repeat in a cycle of length 4. To find the 2023rd derivative, we need to find the remainder of 2023 divided by 4. 2023=4×505+32023 = 4 \times 505 + 3. The remainder is 3. Therefore, the 2023rd derivative is the same as the 3rd derivative. g(2023)(x)=g′′′(x)=−cos⁡(x)−sin⁡(x)g^{(2023)}(x) = g'''(x) = -\cos(x) - \sin(x).

Question 16

Let f(x)=ln⁡(ex2+sin⁡(x))f(x) = \ln(e^{x^2} + \sin(x)). What is the instantaneous rate of change of ff at x=0x=0?

  1. 00
  2. 11 (correct answer)
  3. 22
  4. 1/21/2
Explanation: The instantaneous rate of change of a function is its derivative. We need to find f′(x)f'(x) and evaluate it at x=0x=0. We use the chain rule, where the outer function is ln⁡(u)\ln(u) and the inner function is u=ex2+sin⁡(x)u = e^{x^2} + \sin(x). f′(x)=1ex2+sin⁡(x)⋅ddx(ex2+sin⁡(x))f'(x) = \frac{1}{e^{x^2} + \sin(x)} \cdot \frac{d}{dx}(e^{x^2} + \sin(x)). The derivative of the inner part requires the chain rule for ex2e^{x^2}: ddx(ex2+sin⁡(x))=ex2⋅(2x)+cos⁡(x)\frac{d}{dx}(e^{x^2} + \sin(x)) = e^{x^2} \cdot (2x) + \cos(x). So, f′(x)=2xex2+cos⁡(x)ex2+sin⁡(x)f'(x) = \frac{2xe^{x^2} + \cos(x)}{e^{x^2} + \sin(x)}. Now, we evaluate f′(x)f'(x) at x=0x=0: f′(0)=2(0)e02+cos⁡(0)e02+sin⁡(0)=0⋅1+11+0=11=1f'(0) = \frac{2(0)e^{0^2} + \cos(0)}{e^{0^2} + \sin(0)} = \frac{0 \cdot 1 + 1}{1 + 0} = \frac{1}{1} = 1.

Question 17

Find the equation of the tangent line to the graph of h(x)=e2xsin⁡(x)+1h(x) = \frac{e^{2x}}{\sin(x)+1} at x=0x=0.

  1. y=1y = 1
  2. y=x+1y = x + 1 (correct answer)
  3. y=2x+1y = 2x + 1
  4. y=−x+1y = -x + 1
Explanation: First, find the point of tangency by evaluating h(0)h(0): h(0)=e2(0)sin⁡(0)+1=e00+1=11=1h(0) = \frac{e^{2(0)}}{\sin(0)+1} = \frac{e^0}{0+1} = \frac{1}{1} = 1. The point is (0,1)(0, 1). Next, find the slope of the tangent line by finding the derivative h′(x)h'(x) using the quotient rule: h′(x)=(ddxe2x)(sin⁡(x)+1)−e2x(ddx(sin⁡(x)+1))(sin⁡(x)+1)2=(2e2x)(sin⁡(x)+1)−e2x(cos⁡(x))(sin⁡(x)+1)2h'(x) = \frac{(\frac{d}{dx}e^{2x})(\sin(x)+1) - e^{2x}(\frac{d}{dx}(\sin(x)+1))}{(\sin(x)+1)^2} = \frac{(2e^{2x})(\sin(x)+1) - e^{2x}(\cos(x))}{(\sin(x)+1)^2}. Evaluate the slope at x=0x=0: h′(0)=(2e0)(sin⁡(0)+1)−e0(cos⁡(0))(sin⁡(0)+1)2=2(1)(0+1)−1(1)(0+1)2=2−11=1h'(0) = \frac{(2e^0)(\sin(0)+1) - e^0(\cos(0))}{(\sin(0)+1)^2} = \frac{2(1)(0+1) - 1(1)}{(0+1)^2} = \frac{2-1}{1} = 1. Using the point-slope form, the equation of the tangent line is y−1=1(x−0)y - 1 = 1(x - 0), which simplifies to y=x+1y = x + 1.

Question 18

Given the equation sin⁡(y)+cos⁡(x)=ex+ln⁡(y)\sin(y) + \cos(x) = e^x + \ln(y), find the value of dydx\frac{dy}{dx} at the point (0,1)(0, 1).

  1. 11−cos⁡(1)\frac{1}{1-\cos(1)}
  2. 1cos⁡(1)\frac{1}{\cos(1)}
  3. 1cos⁡(1)+1\frac{1}{\cos(1)+1}
  4. 1cos⁡(1)−1\frac{1}{\cos(1)-1} (correct answer)
Explanation: We use implicit differentiation with respect to xx on both sides of the equation: ddx(sin⁡(y)+cos⁡(x))=ddx(ex+ln⁡(y))\frac{d}{dx}(\sin(y) + \cos(x)) = \frac{d}{dx}(e^x + \ln(y)) cos⁡(y)dydx−sin⁡(x)=ex+1ydydx\cos(y) \frac{dy}{dx} - \sin(x) = e^x + \frac{1}{y} \frac{dy}{dx} Now, we solve for dydx\frac{dy}{dx}. First, group the terms containing dydx\frac{dy}{dx}: cos⁡(y)dydx−1ydydx=ex+sin⁡(x)\cos(y) \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = e^x + \sin(x) Factor out dydx\frac{dy}{dx}: dydx(cos⁡(y)−1y)=ex+sin⁡(x)\frac{dy}{dx} (\cos(y) - \frac{1}{y}) = e^x + \sin(x) dydx=ex+sin⁡(x)cos⁡(y)−1y\frac{dy}{dx} = \frac{e^x + \sin(x)}{\cos(y) - \frac{1}{y}} Finally, substitute the point (x,y)=(0,1)(x, y) = (0, 1) into the expression for dydx\frac{dy}{dx}: dydx∣(0,1)=e0+sin⁡(0)cos⁡(1)−11=1+0cos⁡(1)−1=1cos⁡(1)−1\frac{dy}{dx} \bigg|_{(0,1)} = \frac{e^0 + \sin(0)}{\cos(1) - \frac{1}{1}} = \frac{1 + 0}{\cos(1) - 1} = \frac{1}{\cos(1) - 1}.

Question 19

Let h(x)=f(sin⁡(x))h(x) = f(\sin(x)) and g(x)=ef(x)g(x) = e^{f(x)}. Given that f(0)=ln⁡(3)f(0)=\ln(3) and f′(0)=2f'(0)=2, find the value of h′(0)+g′(0)h'(0) + g'(0).

  1. 22
  2. 55
  3. 66
  4. 88 (correct answer)
Explanation: We need to find the derivatives of h(x)h(x) and g(x)g(x) and evaluate them at x=0x=0. For h(x)=f(sin⁡(x))h(x) = f(\sin(x)), we use the chain rule: h′(x)=f′(sin⁡(x))⋅cos⁡(x)h'(x) = f'(\sin(x)) \cdot \cos(x). Evaluating at x=0x=0: h′(0)=f′(sin⁡(0))⋅cos⁡(0)=f′(0)⋅1=2⋅1=2h'(0) = f'(\sin(0)) \cdot \cos(0) = f'(0) \cdot 1 = 2 \cdot 1 = 2. For g(x)=ef(x)g(x) = e^{f(x)}, we use the chain rule: g′(x)=ef(x)⋅f′(x)g'(x) = e^{f(x)} \cdot f'(x). Evaluating at x=0x=0: g′(0)=ef(0)⋅f′(0)=eln⁡(3)⋅2=3⋅2=6g'(0) = e^{f(0)} \cdot f'(0) = e^{\ln(3)} \cdot 2 = 3 \cdot 2 = 6. Finally, we add the two results: h′(0)+g′(0)=2+6=8h'(0) + g'(0) = 2 + 6 = 8.

Question 20

Let h(x)=ef(x)h(x) = e^{f(x)}, where f(2)=ln⁡3f(2) = \ln 3 and f′(2)=4f'(2) = 4. What is the value of h′(2)h'(2)?

  1. 4
  2. 7
  3. 12 (correct answer)
  4. 4ln⁡34\ln 3
Explanation: Using the chain rule, the derivative of h(x)h(x) is h′(x)=ef(x)⋅f′(x)h'(x) = e^{f(x)} \cdot f'(x). To find h′(2)h'(2), we substitute x=2x=2: h′(2)=ef(2)⋅f′(2)h'(2) = e^{f(2)} \cdot f'(2). We are given f(2)=ln⁡3f(2) = \ln 3 and f′(2)=4f'(2) = 4. Substituting these values gives h′(2)=eln⁡3⋅4=3⋅4=12h'(2) = e^{\ln 3} \cdot 4 = 3 \cdot 4 = 12.