Calculus 1 Quiz: Implicit Differentiation
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Implicit DifferentiationQuestion 1 of 20

If x2+y2=R2x^2 + y^2 = R^2 for a constant RR, which of the following is an expression for d2ydx2\frac{d^2y}{dx^2} in terms of only yy?

R2/y3R^2/y^3
−1/y-1/y
−x/y2-x/y^2
−R2/y3-R^2/y^3
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Calculus 1 Quiz

Calculus 1 Quiz: Implicit Differentiation

Practice Implicit Differentiation 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 Implicit Differentiation, 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

If x2+y2=R2x^2 + y^2 = R^2 for a constant RR, which of the following is an expression for d2ydx2\frac{d^2y}{dx^2} in terms of only yy?

  1. R2/y3R^2/y^3
  2. −1/y-1/y
  3. −x/y2-x/y^2
  4. −R2/y3-R^2/y^3 (correct answer)
Explanation: When you encounter a problem asking for the second derivative of an implicitly defined function, you need to use implicit differentiation twice and then eliminate unwanted variables. Starting with x2+y2=R2x^2 + y^2 = R^2, take the first derivative with respect to xx: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0 Solving for the first derivative: dydx=−xy\frac{dy}{dx} = -\frac{x}{y} Now take the derivative again to find d2ydx2\frac{d^2y}{dx^2}. Using the quotient rule on −xy-\frac{x}{y}: d2ydx2=−y(1)−xdydxy2=−y−x(−xy)y2=−y+x2yy2\frac{d^2y}{dx^2} = -\frac{y(1) - x\frac{dy}{dx}}{y^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} Simplifying: d2ydx2=−y2+x2y3\frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3} Since x2+y2=R2x^2 + y^2 = R^2, we get: d2ydx2=−R2y3\frac{d^2y}{dx^2} = -\frac{R^2}{y^3} Choice A gives R2/y3R^2/y^3 — this misses the negative sign that comes from the implicit differentiation process. Choice B gives −1/y-1/y — this looks like someone only took the first derivative and made errors. Choice C gives −x/y2-x/y^2 — this is close to the first derivative but with the wrong denominator power, and it still contains xx rather than being expressed purely in terms of yy. Remember: when a problem asks for a derivative "in terms of only" one variable, you must substitute back using the original constraint equation to eliminate all other variables from your final answer.

Question 2

The line y=−x+ky = -x + k is tangent to the folium of Descartes, x3+y3−6xy=0x^3 + y^3 - 6xy = 0, at a point in the first quadrant. What is the value of kk?

  1. 66 (correct answer)
  2. 00
  3. −6-6
  4. 33
Explanation: The slope of the tangent line y=−x+ky = -x + k is -1. We need to find the point on the curve where dydx=−1\frac{dy}{dx} = -1. Differentiating the curve's equation: 3x2+3y2dydx−(6y+6xdydx)=03x^2 + 3y^2\frac{dy}{dx} - (6y + 6x\frac{dy}{dx}) = 0. Solving for dydx\frac{dy}{dx} yields dydx=6y−3x23y2−6x=2y−x2y2−2x\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}. Setting the slope to -1: 2y−x2y2−2x=−1\frac{2y - x^2}{y^2 - 2x} = -1, which implies 2y−x2=−y2+2x2y - x^2 = -y^2 + 2x, or y2+2y=x2+2xy^2 + 2y = x^2 + 2x. Completing the square gives (y+1)2=(x+1)2(y+1)^2 = (x+1)^2, so y+1=±(x+1)y+1 = \pm(x+1). This leads to two cases: y=xy=x or y=−x−2y=-x-2. Substituting y=xy=x into the original equation gives 2x3−6x2=02x^3 - 6x^2 = 0, with solutions x=0x=0 (the origin, a singular point) and x=3x=3. The point in the first quadrant is (3,3)(3, 3). Substituting y=−x−2y=-x-2 yields −8=0-8=0, which is impossible. So the point of tangency is (3,3)(3, 3). The tangent line y=−x+ky = -x + k must pass through this point: 3=−3+k3 = -3 + k, which gives k=6k = 6.

Question 3

Consider the curve defined by x3−y3=7x^3 - y^3 = 7. Find the value of the second derivative d2ydx2\frac{d^2y}{dx^2} at the point (2,1)(2, 1).

  1. 44
  2. −28-28 (correct answer)
  3. −12-12
  4. 2828
Explanation: When you encounter implicit differentiation problems asking for the second derivative, you need to find the first derivative, then differentiate it again while carefully applying the chain rule. Starting with x3−y3=7x^3 - y^3 = 7, differentiate both sides with respect to xx: 3x2−3y2dydx=03x^2 - 3y^2\frac{dy}{dx} = 0 Solving for the first derivative: dydx=x2y2\frac{dy}{dx} = \frac{x^2}{y^2} At point (2,1)(2,1): dydx=41=4\frac{dy}{dx} = \frac{4}{1} = 4 For the second derivative, differentiate dydx=x2y2\frac{dy}{dx} = \frac{x^2}{y^2} using the quotient rule: d2ydx2=(y2)(2x)−(x2)(2ydydx)y4=2xy2−2x2ydydxy4\frac{d^2y}{dx^2} = \frac{(y^2)(2x) - (x^2)(2y\frac{dy}{dx})}{y^4} = \frac{2xy^2 - 2x^2y\frac{dy}{dx}}{y^4} Factoring: d2ydx2=2xy(y−xdydx)y4=2x(y−xdydx)y3\frac{d^2y}{dx^2} = \frac{2xy(y - x\frac{dy}{dx})}{y^4} = \frac{2x(y - x\frac{dy}{dx})}{y^3} Substituting our values at (2,1)(2,1) with dydx=4\frac{dy}{dx} = 4: d2ydx2=2(2)(1−2(4))13=4(1−8)1=4(−7)=−28\frac{d^2y}{dx^2} = \frac{2(2)(1 - 2(4))}{1^3} = \frac{4(1-8)}{1} = 4(-7) = -28 The answer is (B) -28. Choice (A) 4 is just the first derivative value. Choice (C) -12 likely comes from arithmetic errors in the quotient rule application. Choice (D) 28 represents a sign error—forgetting that 1−8=−71-8 = -7, not +7+7. Study tip: In second derivative problems with implicit differentiation, organize your work carefully and double-check arithmetic. The quotient rule creates multiple opportunities for sign and calculation errors.

Question 4

Consider the curve defined by the equation x2+y2=25x^2 + y^2 = 25. What is the value of the second derivative, d2ydx2\frac{d^2y}{dx^2}, at the point (3,4)(3, 4)?

  1. −2564-\frac{25}{64} (correct answer)
  2. −34-\frac{3}{4}
  3. −916-\frac{9}{16}
  4. 2564\frac{25}{64}
Explanation: First, differentiate the equation implicitly with respect to xx: 2x+2ydydx=02x + 2y \frac{dy}{dx} = 0, which gives dydx=−xy\frac{dy}{dx} = -\frac{x}{y}. To find the second derivative, differentiate this expression using the quotient rule: d2ydx2=−(1)y−x(dydx)y2\frac{d^2y}{dx^2} = -\frac{(1)y - x(\frac{dy}{dx})}{y^2}. Substitute dydx=−xy\frac{dy}{dx} = -\frac{x}{y} into this equation: d2ydx2=−y−x(−xy)y2=−y+x2yy2=−y2+x2y3\frac{d^2y}{dx^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}. From the original equation, we know x2+y2=25x^2 + y^2 = 25. So, d2ydx2=−25y3\frac{d^2y}{dx^2} = -\frac{25}{y^3}. At the point (3,4)(3, 4), the value is −2543=−2564-\frac{25}{4^3} = -\frac{25}{64}.

Question 5

For the curve defined by x3+y3=1x^3 + y^3 = 1, which of the following is a simplified expression for d2ydx2\frac{d^2y}{dx^2}?

  1. −2xy5-\frac{2x}{y^5} (correct answer)
  2. −2xy4-\frac{2x}{y^4}
  3. 2xy5\frac{2x}{y^5}
  4. −2x4y5-\frac{2x^4}{y^5}
Explanation: First, find dydx\frac{dy}{dx}. Differentiating x3+y3=1x^3 + y^3 = 1 gives 3x2+3y2dydx=03x^2 + 3y^2 \frac{dy}{dx} = 0, so dydx=−x2y2\frac{dy}{dx} = -\frac{x^2}{y^2}. Next, differentiate this expression using the quotient rule to find d2ydx2\frac{d^2y}{dx^2}: d2ydx2=−2x(y2)−x2(2ydydx)(y2)2=−2xy2−2x2ydydxy4\frac{d^2y}{dx^2} = -\frac{2x(y^2) - x^2(2y \frac{dy}{dx})}{(y^2)^2} = -\frac{2xy^2 - 2x^2y \frac{dy}{dx}}{y^4}. Substitute the expression for dydx\frac{dy}{dx}: d2ydx2=−2xy2−2x2y(−x2y2)y4=−2xy2+2x4yy4\frac{d^2y}{dx^2} = -\frac{2xy^2 - 2x^2y(-\frac{x^2}{y^2})}{y^4} = -\frac{2xy^2 + \frac{2x^4}{y}}{y^4}. Multiply the numerator and denominator by yy to simplify: d2ydx2=−2xy3+2x4y5=−2x(y3+x3)y5\frac{d^2y}{dx^2} = -\frac{2xy^3 + 2x^4}{y^5} = -\frac{2x(y^3 + x^3)}{y^5}. Since the original equation is x3+y3=1x^3 + y^3 = 1, we can substitute this in: d2ydx2=−2x(1)y5=−2xy5\frac{d^2y}{dx^2} = -\frac{2x(1)}{y^5} = -\frac{2x}{y^5}.

Question 6

For the curve x=sin⁡(y)x = \sin(y), the tangent line is parallel to 2x−y=52x - y = 5 at a point (x0,y0)(x_0, y_0) where y0y_0 is in the interval (0,π)(0, \pi). What is the value of y0y_0?

  1. π2\frac{\pi}{2}
  2. π6\frac{\pi}{6}
  3. 2π3\frac{2\pi}{3}
  4. π3\frac{\pi}{3} (correct answer)
Explanation: When you encounter a problem asking where a tangent line has a specific slope, you need to find where the derivative equals that target slope. Since this curve is given as x=sin⁡(y)x = \sin(y) rather than the usual y=f(x)y = f(x) form, you'll need to use implicit differentiation or find dxdy\frac{dx}{dy} and then take its reciprocal. The line 2x−y=52x - y = 5 has slope 2, so you need the tangent line to the curve to have slope 2 as well. Starting with x=sin⁡(y)x = \sin(y), differentiate both sides with respect to yy: dxdy=cos⁡(y)\frac{dx}{dy} = \cos(y). Since you want dydx=2\frac{dy}{dx} = 2, and dydx=1dxdy\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}, you need: 1cos⁡(y)=2\frac{1}{\cos(y)} = 2, which means cos⁡(y)=12\cos(y) = \frac{1}{2}. In the interval (0,π)(0, \pi), cosine equals 12\frac{1}{2} when y=π3y = \frac{\pi}{3}. You can verify: when y0=π3y_0 = \frac{\pi}{3}, we have x0=sin⁡(π3)=32x_0 = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, and the slope is indeed 1cos⁡(π/3)=11/2=2\frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2. Looking at the wrong answers: (A) π2\frac{\pi}{2} gives cos⁡(π2)=0\cos\left(\frac{\pi}{2}\right) = 0, making the slope undefined (vertical tangent). (B) π6\frac{\pi}{6} gives cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, so the slope would be 23\frac{2}{\sqrt{3}}, not 2. (C) 2π3\frac{2\pi}{3} gives cos⁡(2π3)=−12\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, making the slope −2-2. Remember: when the curve is given as x=f(y)x = f(y), find dxdy\frac{dx}{dy} first, then take the reciprocal to get dydx\frac{dy}{dx}.

Question 7

For the hyperbola defined by the equation x2−y2+2x+4y=3x^2 - y^2 + 2x + 4y = 3, at what point (x,y)(x,y) is the tangent line vertical?

  1. (1,2)(1, 2)
  2. (−1,0)(-1, 0)
  3. (−1,2)(-1, 2) (correct answer)
  4. (1,0)(1, 0)
Explanation: When you encounter questions about vertical tangent lines, you're dealing with implicit differentiation and the geometric meaning of undefined slope. A tangent line is vertical when its slope is undefined, which occurs when dxdy=0\frac{dx}{dy} = 0 (or equivalently, when dydx\frac{dy}{dx} is undefined). Starting with x2−y2+2x+4y=3x^2 - y^2 + 2x + 4y = 3, use implicit differentiation with respect to xx: 2x−2ydydx+2+4dydx=02x - 2y\frac{dy}{dx} + 2 + 4\frac{dy}{dx} = 0 Solving for dydx\frac{dy}{dx}: 2x+2=2ydydx−4dydx=(2y−4)dydx2x + 2 = 2y\frac{dy}{dx} - 4\frac{dy}{dx} = (2y - 4)\frac{dy}{dx} Therefore: dydx=2x+22y−4=x+1y−2\frac{dy}{dx} = \frac{2x + 2}{2y - 4} = \frac{x + 1}{y - 2} The slope is undefined when the denominator equals zero: y−2=0y - 2 = 0, so y=2y = 2. To find the corresponding xx-value, substitute y=2y = 2 into the original equation: x2−(2)2+2x+4(2)=3x^2 - (2)^2 + 2x + 4(2) = 3 x2−4+2x+8=3x^2 - 4 + 2x + 8 = 3 x2+2x+1=0x^2 + 2x + 1 = 0 (x+1)2=0(x + 1)^2 = 0 x=−1x = -1 The vertical tangent occurs at (−1,2)(-1, 2), which is choice C. Choice A (1,2)(1, 2) gives slope 1+12−2=20\frac{1+1}{2-2} = \frac{2}{0} (undefined), but this point doesn't satisfy the original equation. Choice B (−1,0)(-1, 0) gives slope 0−2=0\frac{0}{-2} = 0 (horizontal tangent). Choice D (1,0)(1, 0) gives slope 2−2=−1\frac{2}{-2} = -1 (neither vertical nor horizontal). Remember: vertical tangents occur when the denominator of dydx\frac{dy}{dx} equals zero, but always verify the point lies on the original curve.

Question 8

What is the equation of the line tangent to the curve cos⁡(x−y)=xey\cos(x-y) = xe^y at the point (0,π2)(0, \frac{\pi}{2})?

  1. y=(1−eπ/2)x+π2y = (1 - e^{\pi/2})x + \frac{\pi}{2} (correct answer)
  2. y=(1+eπ/2)x+π2y = (1 + e^{\pi/2})x + \frac{\pi}{2}
  3. y=(−1−eπ/2)x+π2y = (-1 - e^{\pi/2})x + \frac{\pi}{2}
  4. y=−eπ/2x+π2y = -e^{\pi/2}x + \frac{\pi}{2}
Explanation: Differentiate both sides with respect to xx: −sin⁡(x−y)⋅(1−dydx)=(1)ey+x(eydydx)-\sin(x-y) \cdot (1 - \frac{dy}{dx}) = (1)e^y + x(e^y \frac{dy}{dx}). To find the slope m=dydxm = \frac{dy}{dx} at (0,π2)(0, \frac{\pi}{2}), substitute these values into the differentiated equation: −sin⁡(0−π2)⋅(1−m)=eπ/2+0⋅eπ/2m-\sin(0 - \frac{\pi}{2}) \cdot (1 - m) = e^{\pi/2} + 0 \cdot e^{\pi/2}m. This simplifies to −sin⁡(−π2))(1−m)=eπ/2-\sin(-\frac{\pi}{2}))(1-m) = e^{\pi/2}. Since sin⁡(−π2)=−1\sin(-\frac{\pi}{2}) = -1, we have −(−1))(1−m)=eπ/2-(-1))(1-m) = e^{\pi/2}, which becomes 1−m=eπ/21-m = e^{\pi/2}, so m=1−eπ/2m = 1 - e^{\pi/2}. The equation of the tangent line is y−y1=m(x−x1)y - y_1 = m(x - x_1), which is y−π2=(1−eπ/2)(x−0)y - \frac{\pi}{2} = (1 - e^{\pi/2})(x - 0), or y=(1−eπ/2)x+π2y = (1 - e^{\pi/2})x + \frac{\pi}{2}.

Question 9

What is the yy-intercept of the normal line to the curve x2−xy+y2=3x^2 - xy + y^2 = 3 at the point (−1,1)(-1, 1)?

  1. −1-1
  2. 22
  3. 11
  4. 00 (correct answer)
Explanation: When you encounter a problem asking for the y-intercept of a normal line to a curve, you're dealing with implicit differentiation and perpendicular lines. The normal line is perpendicular to the tangent line at the given point. First, find the slope of the tangent line by implicitly differentiating x2−xy+y2=3x^2 - xy + y^2 = 3. Taking the derivative with respect to xx: 2x−y−xdydx+2ydydx=02x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx}: dydx=2x−yx−2y\frac{dy}{dx} = \frac{2x - y}{x - 2y}. At the point (−1,1)(-1, 1), the slope of the tangent line is dydx=2(−1)−1(−1)−2(1)=−3−3=1\frac{dy}{dx} = \frac{2(-1) - 1}{(-1) - 2(1)} = \frac{-3}{-3} = 1. Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal: mnormal=−1m_{normal} = -1. Using point-slope form with the point (−1,1)(-1, 1) and slope −1-1: y−1=−1(x−(−1))y - 1 = -1(x - (-1)), which simplifies to y−1=−(x+1)y - 1 = -(x + 1), or y=−xy = -x. The y-intercept occurs when x=0x = 0, giving y=0y = 0. Choice A) −1-1 would result from using the x-coordinate as the y-intercept. Choice B) 2 might come from incorrectly calculating the slope or using the wrong point-slope formula. Choice C) 1 would result from using the y-coordinate of the given point as the y-intercept. Remember: for normal line problems, always find the tangent slope first through implicit differentiation, then use the negative reciprocal for the normal slope. The y-intercept is found by setting x=0x = 0 in your final equation.

Question 10

A curve is defined by x3+by3=1x^3 + by^3 = 1, where bb is a non-zero constant. If the point (x0,y0)(x_0, y_0) is on the curve, which of the following expressions represents the slope of the tangent line at that point in terms of x0x_0 and y0y_0 only?

  1. −x02/(by02)-x_0^2 / (by_0^2)
  2. x02y0/(1−x03)x_0^2 y_0 / (1 - x_0^3)
  3. −x02y0/(1−x03)-x_0^2 y_0 / (1 - x_0^3) (correct answer)
  4. −x02y0/(1+x03)-x_0^2 y_0 / (1 + x_0^3)
Explanation: First, we find the derivative dydx\frac{dy}{dx} by implicit differentiation: ddx(x3+by3)=ddx(1)\frac{d}{dx}(x^3 + by^3) = \frac{d}{dx}(1) 3x2+b(3y2)dydx=03x^2 + b(3y^2) \frac{dy}{dx} = 0 Solving for dydx\frac{dy}{dx} gives: 3by2dydx=−3x2  ⟹  dydx=−x2by23by^2 \frac{dy}{dx} = -3x^2 \implies \frac{dy}{dx} = -\frac{x^2}{by^2} This gives the slope at (x0,y0)(x_0, y_0) as m=−x02by02m = -\frac{x_0^2}{by_0^2}. However, the question asks for the slope in terms of x0x_0 and y0y_0 only, so we must eliminate the constant bb. Since the point (x0,y0)(x_0, y_0) is on the curve, it satisfies the original equation: x03+by03=1x_0^3 + by_0^3 = 1. We can solve this equation for bb: by03=1−x03  ⟹  b=1−x03y03by_0^3 = 1 - x_0^3 \implies b = \frac{1 - x_0^3}{y_0^3} Now substitute this expression for bb into the slope formula: m=−x02(1−x03y03)y02=−x021−x03y0=−x02y01−x03m = -\frac{x_0^2}{(\frac{1 - x_0^3}{y_0^3})y_0^2} = -\frac{x_0^2}{\frac{1 - x_0^3}{y_0}} = -\frac{x_0^2 y_0}{1 - x_0^3} Distractor Rationale: (A) This is the correct derivative but fails to complete the second step of the problem, which is to eliminate the constant bb. (B) This results from a sign error in the initial differentiation, leading to a positive slope expression. (D) This would result from an algebraic error when solving for bb, for instance, by03=1+x03by_0^3 = 1 + x_0^3.

Question 11

A student is asked to find dydx\frac{dy}{dx} for the curve x2y=sin⁡(y)x^2y = \sin(y). The student's work is shown below: Line 1: ddx(x2y)=ddx(sin⁡(y))\frac{d}{dx}(x^2y) = \frac{d}{dx}(\sin(y)) Line 2: 2xy=cos⁡(y)dydx2xy = \cos(y) \frac{dy}{dx} Line 3: dydx=2xycos⁡(y)\frac{dy}{dx} = \frac{2xy}{\cos(y)} In which line did the student first make a mistake?

  1. Line 1, the initial setup is incorrect.
  2. Line 3, the algebraic manipulation to isolate dydx\frac{dy}{dx} is incorrect.
  3. Line 2, the chain rule was not applied correctly.
  4. Line 2, the product rule was not applied correctly. (correct answer)
Explanation: When you encounter implicit differentiation problems like this one, you need to carefully apply differentiation rules to both sides of the equation, remembering that yy is a function of xx. Let's trace through the student's work. Line 1 correctly sets up the problem by differentiating both sides of x2y=sin⁡(y)x^2y = \sin(y). Line 3 correctly isolates dydx\frac{dy}{dx} through algebraic manipulation. However, Line 2 contains a critical error. The mistake occurs in differentiating the left side x2yx^2y. Since this is a product of two functions (x2x^2 and yy), you must use the product rule: ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv'. The correct differentiation should be: ddx(x2y)=ddx(x2)⋅y+x2⋅dydx=2xy+x2dydx\frac{d}{dx}(x^2y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{dy}{dx} = 2xy + x^2\frac{dy}{dx} The student wrote 2xy2xy, which only accounts for the first term of the product rule and ignores the second term x2dydxx^2\frac{dy}{dx}. Answer choice A is incorrect because the initial setup is proper. Answer choice B is wrong since the algebraic manipulation in Line 3 is mathematically sound. Answer choice C misidentifies the error—the chain rule was correctly applied to sin⁡(y)\sin(y) on the right side, giving cos⁡(y)dydx\cos(y)\frac{dy}{dx}. Answer choice D correctly identifies that the product rule was improperly applied. Study tip: In implicit differentiation, always check if you need the product rule when differentiating terms with both xx and yy. Write out both terms of the product rule explicitly to avoid missing one.

Question 12

A particle moves along the curve defined by x2y3=8x^2 y^3 = 8. If the x-coordinate is increasing at a rate of 3 units per second, what is the rate of change of the y-coordinate when the particle is at the point (1,2)(1, 2)?

  1. −4-4 units/sec (correct answer)
  2. 44 units/sec
  3. −12-12 units/sec
  4. 1212 units/sec
Explanation: This is a related rates problem that uses implicit differentiation. We are given the equation x2y3=8x^2 y^3 = 8 and dxdt=3\frac{dx}{dt} = 3. We want to find dydt\frac{dy}{dt} at the point (1,2)(1, 2). Differentiate both sides of the equation with respect to time tt, using the product rule on the left side: ddt(x2y3)=ddt(8)\frac{d}{dt}(x^2 y^3) = \frac{d}{dt}(8) ddt(x2)⋅y3+x2⋅ddt(y3)=0\frac{d}{dt}(x^2) \cdot y^3 + x^2 \cdot \frac{d}{dt}(y^3) = 0 (2xdxdt)y3+x2(3y2dydt)=0(2x \frac{dx}{dt}) y^3 + x^2 (3y^2 \frac{dy}{dt}) = 0 Now, substitute the given values: x=1x=1, y=2y=2, and dxdt=3\frac{dx}{dt} = 3. (2(1)(3))(2)3+(1)2(3(2)2dydt)=0(2(1)(3))(2)^3 + (1)^2 (3(2)^2 \frac{dy}{dt}) = 0 (6)(8)+(1)(12dydt)=0(6)(8) + (1)(12 \frac{dy}{dt}) = 0 48+12dydt=048 + 12 \frac{dy}{dt} = 0 12dydt=−4812 \frac{dy}{dt} = -48 dydt=−4\frac{dy}{dt} = -4 The rate of change of the y-coordinate is −4-4 units per second. Distractor Rationale: (B) This results from a sign error in the final step. (C) This comes from an error in applying the product rule, for example, differentiating x2y3x^2 y^3 as (2x)(3y2dydt)(2x)(3y^2 \frac{dy}{dt}). (D) This results from both the product rule error described for (C) and a sign error.

Question 13

Let yy be a function of xx defined implicitly by [f(y)]2=x2y+7[f(y)]^2 = x^2y + 7, where ff is a differentiable function. If f(2)=3f(2)=3 and f′(2)=12f'(2)=\frac{1}{2}, what is the value of dydx\frac{dy}{dx} at the point (1,2)(1, 2)?

  1. 22 (correct answer)
  2. 44
  3. 11
  4. 12\frac{1}{2}
Explanation: First, differentiate the equation [f(y)]2=x2y+7[f(y)]^2 = x^2y + 7 implicitly with respect to xx. Using the chain rule on the left side and the product rule on the right side, we get: 2f(y)⋅f′(y)⋅dydx=(2xy+x2dydx)+02f(y) \cdot f'(y) \cdot \frac{dy}{dx} = (2xy + x^2 \frac{dy}{dx}) + 0. Now, substitute the given values: the point is (x,y)=(1,2)(x,y) = (1,2), and at this point f(2)=3f(2)=3 and f′(2)=12f'(2)=\frac{1}{2}. The equation becomes: 2f(2)⋅f′(2)⋅dydx=2(1)(2)+(1)2dydx2f(2) \cdot f'(2) \cdot \frac{dy}{dx} = 2(1)(2) + (1)^2 \frac{dy}{dx}. Plugging in the function values: 2(3)(12)dydx=4+dydx2(3)(\frac{1}{2}) \frac{dy}{dx} = 4 + \frac{dy}{dx}. This simplifies to 3dydx=4+dydx3\frac{dy}{dx} = 4 + \frac{dy}{dx}. Solving for dydx\frac{dy}{dx} gives 2dydx=42\frac{dy}{dx} = 4, so dydx=2\frac{dy}{dx} = 2.

Question 14

The curve y=x22y = \frac{x^2}{2} and the circle x2+y2=8x^2+y^2=8 intersect at two points. What is the slope of the line tangent to the circle at the intersection point with a positive xx-coordinate?

  1. 11
  2. 22
  3. −1-1 (correct answer)
  4. −12-\frac{1}{2}
Explanation: When you encounter problems involving tangent lines to curves at intersection points, you need to find where the curves meet, then use implicit differentiation or the derivative to find the slope at that specific point. First, let's find the intersection points. Since both curves pass through the same points, we can substitute y=x22y = \frac{x^2}{2} into the circle equation: x2+(x22)2=8x^2 + \left(\frac{x^2}{2}\right)^2 = 8. This gives us x2+x44=8x^2 + \frac{x^4}{4} = 8, or x4+4x2−32=0x^4 + 4x^2 - 32 = 0. Letting u=x2u = x^2, we get u2+4u−32=0u^2 + 4u - 32 = 0, which factors as (u+8)(u−4)=0(u + 8)(u - 4) = 0. Since u=x2≥0u = x^2 \geq 0, we have u=4u = 4, so x=±2x = \pm 2. The intersection point with positive x-coordinate is (2,2)(2, 2). To find the tangent slope to the circle at this point, use implicit differentiation on x2+y2=8x^2 + y^2 = 8: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−xy\frac{dy}{dx} = -\frac{x}{y}. At point (2,2)(2, 2), the slope is −22=−1-\frac{2}{2} = -1. Looking at the wrong answers: A) 11 is the negative reciprocal of the correct answer—you might get this if you confused the relationship between perpendicular lines. B) 22 could result from forgetting the negative sign or miscalculating xy\frac{x}{y}. D) −12-\frac{1}{2} might come from incorrectly applying the quotient rule or mixing up xx and yy values. Remember: for implicit differentiation of circles, the slope formula dydx=−xy\frac{dy}{dx} = -\frac{x}{y} is your go-to tool. Always substitute the specific point coordinates carefully.

Question 15

Consider the curve defined by the equation x2+y2=25x^2 + y^2 = 25. What is the value of the second derivative, d2y/dx2d^2y/dx^2, at the point (3,4)(3, 4)?

  1. 3/43/4
  2. −25/64-25/64 (correct answer)
  3. 25/6425/64
  4. −25/16-25/16
Explanation: First, differentiate the equation implicitly with respect to xx: 2x+2ydydx=02x + 2y \frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx} gives dydx=−xy\frac{dy}{dx} = -\frac{x}{y}. Next, differentiate this expression with respect to xx using the quotient rule to find the second derivative: d2ydx2=−ddx(x)⋅y−(−x)⋅ddx(y)y2=−1⋅y+x⋅dydxy2\frac{d^2y}{dx^2} = \frac{-\frac{d}{dx}(x) \cdot y - (-x) \cdot \frac{d}{dx}(y)}{y^2} = \frac{-1 \cdot y + x \cdot \frac{dy}{dx}}{y^2} Substitute the expression for dydx\frac{dy}{dx}: d2ydx2=−y+x(−xy)y2=−y−x2yy2=−y2−x2y3\frac{d^2y}{dx^2} = \frac{-y + x(-\frac{x}{y})}{y^2} = \frac{-y - \frac{x^2}{y}}{y^2} = \frac{-y^2 - x^2}{y^3} From the original equation, we know x2+y2=25x^2 + y^2 = 25, so we can substitute this into the numerator: d2ydx2=−(x2+y2)y3=−25y3\frac{d^2y}{dx^2} = -\frac{(x^2 + y^2)}{y^3} = -\frac{25}{y^3} Finally, evaluate at the point (3,4)(3, 4): d2ydx2=−2543=−2564\frac{d^2y}{dx^2} = -\frac{25}{4^3} = -\frac{25}{64} Distractor Rationale: (A) This is the value of the first derivative dydx\frac{dy}{dx} at (−3,4)(-3, 4), not the second derivative at (3,4)(3,4). (C) This results from a sign error in the quotient rule. (D) This results from an algebraic error in simplification, calculating the denominator as y2y^2 instead of y3y^3.

Question 16

What is the equation of the tangent line to the curve defined by y2+cos⁡(xy)=x2y^2 + \cos(xy) = x^2 at the point (1,0)(1, 0)?

  1. y=2x−2y = 2x - 2
  2. y=0y = 0
  3. x=1x = 1 (correct answer)
  4. y=−2x+2y = -2x + 2
Explanation: First, we find the slope of the tangent line by differentiating the equation implicitly with respect to xx. We must use the chain rule for the cos⁡(xy)\cos(xy) term, which also requires the product rule for its argument xyxy. ddx(y2+cos⁡(xy))=ddx(x2)\frac{d}{dx}(y^2 + \cos(xy)) = \frac{d}{dx}(x^2) 2ydydx−sin⁡(xy)⋅(1⋅y+xdydx)=2x2y\frac{dy}{dx} - \sin(xy) \cdot \left(1 \cdot y + x \frac{dy}{dx}\right) = 2x Distribute the −sin⁡(xy)-\sin(xy) term: 2ydydx−ysin⁡(xy)−xsin⁡(xy)dydx=2x2y\frac{dy}{dx} - y\sin(xy) - x\sin(xy)\frac{dy}{dx} = 2x Group the terms with dydx\frac{dy}{dx} and solve for it: dydx(2y−xsin⁡(xy))=2x+ysin⁡(xy)\frac{dy}{dx}(2y - x\sin(xy)) = 2x + y\sin(xy) dydx=2x+ysin⁡(xy)2y−xsin⁡(xy)\frac{dy}{dx} = \frac{2x + y\sin(xy)}{2y - x\sin(xy)} Now, evaluate the slope at the point (1,0)(1, 0): m=2(1)+0⋅sin⁡(0)2(0)−1⋅sin⁡(0)=20m = \frac{2(1) + 0 \cdot \sin(0)}{2(0) - 1 \cdot \sin(0)} = \frac{2}{0} The slope is undefined, which indicates a vertical tangent line. The equation of a vertical line passing through the point (1,0)(1, 0) is x=1x = 1. Distractor Rationale: (A) This line has a slope of 22. This slope would be found if the derivative was mistakenly calculated as dydx=2x\frac{dy}{dx} = 2x. (B) This is the equation for a horizontal tangent line, which would occur if the slope was 00. (D) This line has a slope of −2-2, resulting from potential sign errors during differentiation or algebraic manipulation.

Question 17

For the curve defined by x2+xy+y2=3x^2 + xy + y^2 = 3, what are the x-coordinates of all points where the tangent line is vertical?

  1. x=1x=1 and x=−1x=-1
  2. x=2x=2 and x=−2x=-2 (correct answer)
  3. x=3x=\sqrt{3} and x=−3x=-\sqrt{3}
  4. x=1/3x=1/\sqrt{3} and x=−1/3x=-1/\sqrt{3}
Explanation: First, find the derivative dydx\frac{dy}{dx} by implicit differentiation: ddx(x2+xy+y2)=ddx(3)\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(3) 2x+(1⋅y+xdydx)+2ydydx=02x + \left(1 \cdot y + x \frac{dy}{dx}\right) + 2y \frac{dy}{dx} = 0 Isolate dydx\frac{dy}{dx}: dydx(x+2y)=−2x−y\frac{dy}{dx}(x + 2y) = -2x - y dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x+y}{x+2y} A vertical tangent occurs when the slope is undefined, which means the denominator of dydx\frac{dy}{dx} is zero (and the numerator is non-zero). Set the denominator to zero: x+2y=0x + 2y = 0, which implies x=−2yx = -2y. Substitute x=−2yx = -2y into the original equation of the curve to find the points: (−2y)2+(−2y)y+y2=3(-2y)^2 + (-2y)y + y^2 = 3 4y2−2y2+y2=34y^2 - 2y^2 + y^2 = 3 3y2=33y^2 = 3 y2=1  ⟹  y=±1y^2 = 1 \implies y = \pm 1 Now find the corresponding x-coordinates using x=−2yx = -2y: If y=1y=1, then x=−2(1)=−2x = -2(1) = -2. If y=−1y=-1, then x=−2(−1)=2x = -2(-1) = 2. The x-coordinates are x=2x=2 and x=−2x=-2. Distractor Rationale: (A) These are the x-coordinates for the points where the tangent line is horizontal (numerator of dydx\frac{dy}{dx} is zero). (C) These are the x-intercepts of the curve, found by setting y=0y=0 in the original equation. (D) These x-coordinates arise if an algebraic error is made, such as setting y=xy=x and solving.

Question 18

At which point in the first quadrant is the tangent line to the curve x2−xy+y2=1x^2 - xy + y^2 = 1 parallel to the line y=−x+5y = -x + 5?

  1. (1,1)(1, 1) (correct answer)
  2. (−1,−1)(-1, -1)
  3. (1/3,2/3)(1/\sqrt{3}, 2/\sqrt{3})
  4. (1,0)(1, 0)
Explanation: The line y=−x+5y = -x + 5 has a slope of m=−1m = -1. We need to find the point (x,y)(x, y) on the curve where the slope of the tangent line is also −1-1. First, find the derivative dydx\frac{dy}{dx} using implicit differentiation: ddx(x2−xy+y2)=ddx(1)\frac{d}{dx}(x^2 - xy + y^2) = \frac{d}{dx}(1) 2x−(1⋅y+xdydx)+2ydydx=02x - \left(1 \cdot y + x \frac{dy}{dx}\right) + 2y \frac{dy}{dx} = 0 Isolate dydx\frac{dy}{dx}: 2x−y=dydx(x−2y)  ⟹  dydx=2x−yx−2y2x - y = \frac{dy}{dx}(x - 2y) \implies \frac{dy}{dx} = \frac{2x - y}{x - 2y} Set the slope equal to −1-1: −1=2x−yx−2y-1 = \frac{2x - y}{x - 2y} −(x−2y)=2x−y  ⟹  −x+2y=2x−y  ⟹  3y=3x  ⟹  y=x-(x - 2y) = 2x - y \implies -x + 2y = 2x - y \implies 3y = 3x \implies y = x Now substitute y=xy=x into the original equation of the curve to find the points: x2−x(x)+(x)2=1x^2 - x(x) + (x)^2 = 1 x2−x2+x2=1  ⟹  x2=1  ⟹  x=±1x^2 - x^2 + x^2 = 1 \implies x^2 = 1 \implies x = \pm 1 Since y=xy = x, the points are (1,1)(1, 1) and (−1,−1)(-1, -1). The question asks for the point in the first quadrant, which is (1,1)(1, 1). Distractor Rationale: (B) This is the other point on the curve with a slope of −1-1, but it lies in the third quadrant. (C) This is a point on the curve where the tangent line is horizontal (slope is 0). (D) This is a point on the curve, but the slope at this point is 2(1)−01−2(0)=2\frac{2(1)-0}{1-2(0)} = 2, not −1-1.

Question 19

For the curve defined by the equation exy=x−ye^{xy} = x - y, what is the slope of the tangent line at the point (1,0)(1, 0)?

  1. 11
  2. 00
  3. −1/2-1/2
  4. 1/21/2 (correct answer)
Explanation: To find the slope, we differentiate the equation implicitly with respect to xx. This requires the chain rule on the left side and the product rule for the exponent xyxy. ddx(exy)=ddx(x−y)\frac{d}{dx}(e^{xy}) = \frac{d}{dx}(x - y) exy⋅ddx(xy)=1−dydxe^{xy} \cdot \frac{d}{dx}(xy) = 1 - \frac{dy}{dx} exy⋅(1⋅y+xdydx)=1−dydxe^{xy} \cdot \left(1 \cdot y + x \frac{dy}{dx}\right) = 1 - \frac{dy}{dx} Distribute the exye^{xy} term: yexy+xexydydx=1−dydxy e^{xy} + x e^{xy} \frac{dy}{dx} = 1 - \frac{dy}{dx} Now, group all terms with dydx\frac{dy}{dx} on one side and solve: xexydydx+dydx=1−yexyx e^{xy} \frac{dy}{dx} + \frac{dy}{dx} = 1 - y e^{xy} dydx(xexy+1)=1−yexy\frac{dy}{dx} (x e^{xy} + 1) = 1 - y e^{xy} dydx=1−yexyxexy+1\frac{dy}{dx} = \frac{1 - y e^{xy}}{x e^{xy} + 1} Finally, substitute the coordinates of the point (1,0)(1, 0) into this expression: m=1−0⋅e1⋅01⋅e1⋅0+1=1−01⋅1+1=12m = \frac{1 - 0 \cdot e^{1 \cdot 0}}{1 \cdot e^{1 \cdot 0} + 1} = \frac{1 - 0}{1 \cdot 1 + 1} = \frac{1}{2} Distractor Rationale: (A) This results from forgetting to apply the chain rule to the −y-y term on the right-hand side, treating its derivative as 00 instead of dydx\frac{dy}{dx}. (B) This results from forgetting the product rule for the term xyxy in the exponent, treating its derivative as just yy or xx. (C) This results from a sign error when isolating the dydx\frac{dy}{dx} term, for example, getting yexy−1xexy+1\frac{y e^{xy} - 1}{x e^{xy} + 1}.

Question 20

For the curve defined by x2+xy+y2=3x^2 + xy + y^2 = 3, what are the y-coordinates of all points where the tangent line is horizontal?

  1. y=1y=1 and y=−1y=-1
  2. y=2y=2 and y=−2y=-2 (correct answer)
  3. y=3y=\sqrt{3} and y=−3y=-\sqrt{3}
  4. y=1/3y=1/\sqrt{3} and y=−1/3y=-1/\sqrt{3}
Explanation: First, find the derivative dydx\frac{dy}{dx} by implicit differentiation: ddx(x2+xy+y2)=ddx(3)\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(3) 2x+y+xdydx+2ydydx=02x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 Isolate dydx\frac{dy}{dx}: dydx(x+2y)=−2x−y\frac{dy}{dx}(x + 2y) = -2x - y dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x+y}{x+2y} A horizontal tangent occurs when the slope is zero, which means the numerator of dydx\frac{dy}{dx} is zero. Set the numerator to zero: 2x+y=02x + y = 0, which implies y=−2xy = -2x. Substitute y=−2xy = -2x into the original equation of the curve to find the points: x2+x(−2x)+(−2x)2=3x^2 + x(-2x) + (-2x)^2 = 3 x2−2x2+4x2=3x^2 - 2x^2 + 4x^2 = 3 3x2=33x^2 = 3 x2=1  ⟹  x=±1x^2 = 1 \implies x = \pm 1 Now find the corresponding y-coordinates using y=−2xy = -2x: If x=1x=1, then y=−2(1)=−2y = -2(1) = -2. If x=−1x=-1, then y=−2(−1)=2y = -2(-1) = 2. The y-coordinates are y=2y=2 and y=−2y=-2. Distractor Rationale: (A) These are the y-coordinates for the points where the tangent line is vertical (denominator of dydx\frac{dy}{dx} is zero). (C) These are the y-intercepts of the curve, found by setting x=0x=0 in the original equation. (D) This is an incorrect result possibly arising from a calculation error, such as setting y=xy=x.