Business Calculus Quiz: Implicit Differentiation
14 questions · exam conditions
0:00
Implicit DifferentiationQuestion 1 of 14

Consider the curve x3+y3−6x2=0x^3 + y^3 - 6x^2 = 0. What is the slope of the normal line to the curve at the point where x=3x=3 and y>0y > 0?

1/31/3
33
-3
-1/3
← Back to quizzes

Business Calculus Quiz

Business Calculus Quiz: Implicit Differentiation

Practice Implicit Differentiation in Business Calculus 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 Business Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Consider the curve x3+y3−6x2=0x^3 + y^3 - 6x^2 = 0. What is the slope of the normal line to the curve at the point where x=3x=3 and y>0y > 0?

  1. 1/31/3
  2. 33
  3. -3 (correct answer)
  4. -1/3
Explanation: First, find the yy-coordinate of the point by substituting x=3x=3 into the equation: (3)3+y3−6(3)2=0  ⟹  27+y3−54=0  ⟹  y3=27  ⟹  y=3(3)^3 + y^3 - 6(3)^2 = 0 \implies 27 + y^3 - 54 = 0 \implies y^3 = 27 \implies y=3. The point is (3,3)(3, 3). Next, find the slope of the tangent line, dy/dxdy/dx, by implicit differentiation: 3x2+3y2dydx−12x=03x^2 + 3y^2\frac{dy}{dx} - 12x = 0. 3y2dydx=12x−3x23y^2\frac{dy}{dx} = 12x - 3x^2. dydx=12x−3x23y2=4x−x2y2\frac{dy}{dx} = \frac{12x - 3x^2}{3y^2} = \frac{4x - x^2}{y^2}. Evaluate the slope of the tangent at (3,3)(3, 3): mtan=4(3)−(3)2(3)2=12−99=39=13m_{tan} = \frac{4(3) - (3)^2}{(3)^2} = \frac{12 - 9}{9} = \frac{3}{9} = \frac{1}{3}. The slope of the normal line is the negative reciprocal of the tangent slope: mnormal=−1mtan=−11/3=−3m_{normal} = -\frac{1}{m_{tan}} = -\frac{1}{1/3} = -3.

Question 2

The production of a firm is modeled by an isoquant equation x2/3+y2/3=13x^{2/3} + y^{2/3} = 13, where xx is units of capital and yy is units of labor. Find the marginal rate of technical substitution, defined as −dy/dx-dy/dx, at the production level (27,8)(27, 8).

  1. −2/3-2/3
  2. 2/32/3 (correct answer)
  3. −3/2-3/2
  4. 3/23/2
Explanation: First, we find dy/dxdy/dx by implicit differentiation. Differentiating x2/3+y2/3=13x^{2/3} + y^{2/3} = 13 with respect to xx: 23x−1/3+23y−1/3dydx=0\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0. Solving for dy/dxdy/dx: 23y−1/3dydx=−23x−1/3\frac{2}{3}y^{-1/3}\frac{dy}{dx} = -\frac{2}{3}x^{-1/3}. dydx=−x−1/3y−1/3=−(yx)1/3\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3}. Now, evaluate dy/dxdy/dx at the point (27,8)(27, 8): dydx=−(827)1/3=−23\frac{dy}{dx} = -\left(\frac{8}{27}\right)^{1/3} = -\frac{2}{3}. The marginal rate of technical substitution (MRTS) is defined as −dy/dx-dy/dx. Therefore, MRTS =−(−23)=23= -(-\frac{2}{3}) = \frac{2}{3}.

Question 3

For the ellipse described by x2+4y2=16x^2 + 4y^2 = 16, find the rate of change of xx with respect to yy at the point (2,3)(2, \sqrt{3}).

  1. −36-\frac{\sqrt{3}}{6}
  2. 36\frac{\sqrt{3}}{6}
  3. −23-2\sqrt{3} (correct answer)
  4. 232\sqrt{3}
Explanation: The 'rate of change of xx with respect to yy' is dx/dydx/dy. We can find this by differentiating the equation with respect to yy. 2xdxdy+8y=02x\frac{dx}{dy} + 8y = 0. Solving for dx/dydx/dy: 2xdxdy=−8y2x\frac{dx}{dy} = -8y. dxdy=−8y2x=−4yx\frac{dx}{dy} = -\frac{8y}{2x} = -\frac{4y}{x}. Now, we evaluate this derivative at the point (2,3)(2, \sqrt{3}): dxdy=−4(3)2=−23\frac{dx}{dy} = -\frac{4(\sqrt{3})}{2} = -2\sqrt{3}. Alternatively, one could find dy/dxdy/dx and then take the reciprocal, since dx/dy=1/(dy/dx)dx/dy = 1/(dy/dx).

Question 4

A company's production of two goods, xx and yy, is constrained by the equation x2+2y2−8x=20x^2 + 2y^2 - 8x = 20. At what positive value of xx is the tangent to this production curve horizontal?

  1. x=10x = 10
  2. x=4x = 4 (correct answer)
  3. x=32x = 3\sqrt{2}
  4. x=8x = 8
Explanation: A horizontal tangent occurs where the slope, dy/dxdy/dx, is equal to zero. First, we find dy/dxdy/dx by implicit differentiation: 2x+4ydydx−8=02x + 4y\frac{dy}{dx} - 8 = 0. Solving for dy/dxdy/dx: 4ydydx=8−2x4y\frac{dy}{dx} = 8 - 2x, so dydx=8−2x4y=4−x2y\frac{dy}{dx} = \frac{8 - 2x}{4y} = \frac{4 - x}{2y}. For the tangent line to be horizontal, the slope must be zero, which means the numerator of the derivative must be zero: 4−x=04 - x = 0. This gives x=4x = 4.

Question 5

Find the slope of the tangent line to the curve defined by the equation x3+y3=9xyx^3 + y^3 = 9xy at the point (2,4)(2, 4).

  1. 4/54/5 (correct answer)
  2. 5/45/4
  3. 1/21/2
  4. 4/34/3
Explanation: To find the slope of the tangent line, we need to find the derivative dy/dxdy/dx using implicit differentiation. Differentiating both sides of the equation with respect to xx gives: 3x2+3y2dydx=9(1⋅y+x⋅dydx)3x^2 + 3y^2 \frac{dy}{dx} = 9(1 \cdot y + x \cdot \frac{dy}{dx}). Using the product rule on the right side. Now, we solve for dy/dxdy/dx: 3x2+3y2dydx=9y+9xdydx3x^2 + 3y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}. 3y2dydx−9xdydx=9y−3x23y^2 \frac{dy}{dx} - 9x \frac{dy}{dx} = 9y - 3x^2. dydx(3y2−9x)=9y−3x2\frac{dy}{dx}(3y^2 - 9x) = 9y - 3x^2. dydx=9y−3x23y2−9x=3y−x2y2−3x\frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x} = \frac{3y - x^2}{y^2 - 3x}. Finally, substitute the point (2,4)(2, 4) into the derivative: dydx=3(4)−(2)2(4)2−3(2)=12−416−6=810=45\frac{dy}{dx} = \frac{3(4) - (2)^2}{(4)^2 - 3(2)} = \frac{12 - 4}{16 - 6} = \frac{8}{10} = \frac{4}{5}.

Question 6

The equation x3+y3−3axy=0x^3 + y^3 - 3axy = 0 (where a>0a > 0 is a constant) defines a curve called a folium. If the curve passes through the point (3,3)(3, 3), then dydx\frac{dy}{dx} at this point equals:

  1. dydx=−1\frac{dy}{dx} = -1 (correct answer)
  2. dydx=1\frac{dy}{dx} = 1
  3. dydx=−13\frac{dy}{dx} = -\frac{1}{3}
  4. dydx=13\frac{dy}{dx} = \frac{1}{3}
Explanation: First, find the value of aa using the condition that the curve passes through (3,3)(3,3): 33+33−3a(3)(3)=03^3 + 3^3 - 3a(3)(3) = 0, so 27+27−27a=027 + 27 - 27a = 0, giving 54=27a54 = 27a and a=2a = 2. The equation becomes x3+y3−6xy=0x^3 + y^3 - 6xy = 0. Now differentiate implicitly: 3x2+3y2dydx−6y−6xdydx=03x^2 + 3y^2\frac{dy}{dx} - 6y - 6x\frac{dy}{dx} = 0. Rearranging: (3y2−6x)dydx=6y−3x2(3y^2 - 6x)\frac{dy}{dx} = 6y - 3x^2, so dydx=6y−3x23y2−6x=2y−x2y2−2x\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}. At the point (3,3)(3,3): dydx=2(3)−3232−2(3)=6−99−6=−33=−1\frac{dy}{dx} = \frac{2(3) - 3^2}{3^2 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1. Choice B forgets the negative sign. Choice C results from using −39\frac{-3}{9} instead of −33\frac{-3}{3}. Choice D makes both the sign error and the denominator error.

Question 7

For the curve defined by exy=x−ye^{xy} = x - y, find the value of dy/dxdy/dx at the point (1,0)(1, 0).

  1. 1/21/2 (correct answer)
  2. −1/2-1/2
  3. 00
  4. The derivative is undefined.
Explanation: We use implicit differentiation. The derivative of the left side requires the chain rule and the product rule: exy⋅ddx(xy)=exy(y+xdydx)e^{xy} \cdot \frac{d}{dx}(xy) = e^{xy}(y + x\frac{dy}{dx}). The derivative of the right side is 1−dydx1 - \frac{dy}{dx}. Setting them equal: exy(y+xdydx)=1−dydxe^{xy}(y + x\frac{dy}{dx}) = 1 - \frac{dy}{dx}. yexy+xexydydx=1−dydxye^{xy} + xe^{xy}\frac{dy}{dx} = 1 - \frac{dy}{dx}. Now, group the dy/dxdy/dx terms: xexydydx+dydx=1−yexyxe^{xy}\frac{dy}{dx} + \frac{dy}{dx} = 1 - ye^{xy}. Factor out dy/dxdy/dx: dydx(xexy+1)=1−yexy\frac{dy}{dx}(xe^{xy} + 1) = 1 - ye^{xy}. So, dydx=1−yexyxexy+1\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} + 1}. Substitute the point (1,0)(1, 0): dydx=1−(0)e1⋅0(1)e1⋅0+1=1−01⋅e0+1=11+1=12\frac{dy}{dx} = \frac{1 - (0)e^{1 \cdot 0}}{(1)e^{1 \cdot 0} + 1} = \frac{1 - 0}{1 \cdot e^0 + 1} = \frac{1}{1 + 1} = \frac{1}{2}.

Question 8

Find the slope of the tangent line to the curve y4+x2y2=5y^4 + x^2y^2 = 5 at the point on the curve where y=1y=1 and x>0x > 0.

  1. −1/3-1/3 (correct answer)
  2. 1/31/3
  3. −3-3
  4. 00
Explanation: First, find the xx-coordinate. Substitute y=1y=1 into the equation: (1)4+x2(1)2=5  ⟹  1+x2=5  ⟹  x2=4(1)^4 + x^2(1)^2 = 5 \implies 1 + x^2 = 5 \implies x^2 = 4. Since x>0x > 0, we have x=2x=2. The point is (2,1)(2, 1). Next, differentiate the equation implicitly with respect to xx: 4y3dydx+(2xy2+x2(2ydydx))=04y^3\frac{dy}{dx} + (2xy^2 + x^2(2y\frac{dy}{dx})) = 0. 4y3dydx+2xy2+2x2ydydx=04y^3\frac{dy}{dx} + 2xy^2 + 2x^2y\frac{dy}{dx} = 0. Group terms with dy/dxdy/dx: dydx(4y3+2x2y)=−2xy2\frac{dy}{dx}(4y^3 + 2x^2y) = -2xy^2. dydx=−2xy24y3+2x2y\frac{dy}{dx} = \frac{-2xy^2}{4y^3 + 2x^2y}. Substitute the point (2,1)(2, 1): dydx=−2(2)(1)24(1)3+2(2)2(1)=−44+8=−412=−13\frac{dy}{dx} = \frac{-2(2)(1)^2}{4(1)^3 + 2(2)^2(1)} = \frac{-4}{4 + 8} = \frac{-4}{12} = -\frac{1}{3}.

Question 9

Given the relation (x+y)2=3xy+7(x+y)^2 = 3xy + 7, find the value of dydx\frac{dy}{dx} at the point (3,1)(3, 1).

  1. 5/95/9
  2. −8/5-8/5
  3. 5/175/17
  4. 55 (correct answer)
Explanation: Differentiate both sides with respect to xx. Use the chain rule on the left side and the product rule on the right side: 2(x+y)(1+dydx)=3(1⋅y+x⋅dydx)2(x+y)(1 + \frac{dy}{dx}) = 3(1 \cdot y + x \cdot \frac{dy}{dx}). Now, substitute the coordinates of the point (3,1)(3, 1) into the differentiated equation before solving for dy/dxdy/dx to simplify the algebra. 2(3+1)(1+dydx)=3(1+3dydx)2(3+1)(1 + \frac{dy}{dx}) = 3(1 + 3\frac{dy}{dx}). 2(4)(1+dydx)=3+9dydx2(4)(1 + \frac{dy}{dx}) = 3 + 9\frac{dy}{dx}. 8(1+dydx)=3+9dydx8(1 + \frac{dy}{dx}) = 3 + 9\frac{dy}{dx}. 8+8dydx=3+9dydx8 + 8\frac{dy}{dx} = 3 + 9\frac{dy}{dx}. 8−3=9dydx−8dydx8 - 3 = 9\frac{dy}{dx} - 8\frac{dy}{dx}. 5=dydx5 = \frac{dy}{dx}.

Question 10

Find the equation of the tangent line to the curve defined by x2y−xy2=2x^2y - xy^2 = 2 at the point (2,1)(2, 1).

  1. y=−38x+74y = -\frac{3}{8}x + \frac{7}{4}
  2. x=2x = 2 (correct answer)
  3. y=1y = 1
  4. y=58x−14y = \frac{5}{8}x - \frac{1}{4}
Explanation: First, use implicit differentiation to find the slope dy/dxdy/dx. Applying the product rule to both terms: (2xy+x2dydx)−(y2+x(2ydydx))=0(2xy + x^2\frac{dy}{dx}) - (y^2 + x(2y\frac{dy}{dx})) = 0. 2xy+x2dydx−y2−2xydydx=02xy + x^2\frac{dy}{dx} - y^2 - 2xy\frac{dy}{dx} = 0. Isolate terms with dy/dxdy/dx: dydx(x2−2xy)=y2−2xy\frac{dy}{dx}(x^2 - 2xy) = y^2 - 2xy. dydx=y2−2xyx2−2xy\frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy}. Now, evaluate the slope at (2,1)(2, 1): m=(1)2−2(2)(1)(2)2−2(2)(1)=1−44−4=−30m = \frac{(1)^2 - 2(2)(1)}{(2)^2 - 2(2)(1)} = \frac{1 - 4}{4 - 4} = \frac{-3}{0}. Since the slope is undefined, the tangent line is a vertical line. The equation of a vertical line passing through the point (2,1)(2, 1) is x=2x = 2.

Question 11

Find the derivative dydx\frac{dy}{dx} for the implicit relation x2y=1+y2xx^2y = 1 + y^2x.

  1. y2x2\frac{y^2}{x^2}
  2. y2−2xyx2−2xy\frac{y^2 - 2xy}{x^2 - 2xy} (correct answer)
  3. 2xy2y−x2\frac{2xy}{2y - x^2}
  4. y22x−2xy\frac{y^2}{2x - 2xy}
Explanation: We differentiate both sides of the equation with respect to xx, using the product rule on both x2yx^2y and y2xy^2x. For the left side: ddx(x2y)=2xy+x2dydx\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}. For the right side: ddx(1+y2x)=0+(2ydydx⋅x+y2⋅1)\frac{d}{dx}(1 + y^2x) = 0 + (2y\frac{dy}{dx} \cdot x + y^2 \cdot 1). Setting them equal: 2xy+x2dydx=2xydydx+y22xy + x^2\frac{dy}{dx} = 2xy\frac{dy}{dx} + y^2. Now, we must algebraically solve for dy/dxdy/dx. Group all terms containing dy/dxdy/dx on one side: x2dydx−2xydydx=y2−2xyx^2\frac{dy}{dx} - 2xy\frac{dy}{dx} = y^2 - 2xy. Factor out dy/dxdy/dx: dydx(x2−2xy)=y2−2xy\frac{dy}{dx}(x^2 - 2xy) = y^2 - 2xy. Finally, divide to isolate dy/dxdy/dx: dydx=y2−2xyx2−2xy\frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy}.

Question 12

At which yy-coordinates does the curve defined by x2−xy+y2=3x^2 - xy + y^2 = 3 have a vertical tangent line?

  1. y=1y = 1 and y=−1y = -1 (correct answer)
  2. y=2y = 2 and y=−2y = -2
  3. y=3y = \sqrt{3} and y=−3y = -\sqrt{3}
  4. y=0y = 0
Explanation: A vertical tangent line occurs where the slope dy/dxdy/dx is undefined, meaning the denominator of the derivative is zero. First, differentiate implicitly: 2x−(1⋅y+xdydx)+2ydydx=02x - (1 \cdot y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0. 2x−y−xdydx+2ydydx=02x - y - x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0. dydx(2y−x)=y−2x\frac{dy}{dx}(2y - x) = y - 2x. dydx=y−2x2y−x\frac{dy}{dx} = \frac{y - 2x}{2y - x}. The denominator is zero when 2y−x=02y - x = 0, or x=2yx = 2y. To find the points on the curve where this occurs, substitute x=2yx = 2y back into the original equation: (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, so y2=1y^2 = 1. The yy-coordinates are therefore y=1y = 1 and y=−1y = -1.

Question 13

For the equation x2+xy−3y2=7x^2 + xy - 3y^2 = 7, find dydx\frac{dy}{dx} when x=4x = 4 and y=1y = 1.

  1. dydx=92\frac{dy}{dx} = \frac{9}{2} (correct answer)
  2. dydx=−92\frac{dy}{dx} = -\frac{9}{2}
  3. dydx=52\frac{dy}{dx} = \frac{5}{2}
  4. dydx=−52\frac{dy}{dx} = -\frac{5}{2}
Explanation: Using implicit differentiation: ddx[x2+xy−3y2]=ddx[7]\frac{d}{dx}[x^2 + xy - 3y^2] = \frac{d}{dx}[7]. This gives 2x+y+xdydx−6ydydx=02x + y + x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx}: (x−6y)dydx=−2x−y(x - 6y)\frac{dy}{dx} = -2x - y, so dydx=−2x−yx−6y\frac{dy}{dx} = \frac{-2x - y}{x - 6y}. At (4,1)(4,1): dydx=−8−14−6=−9−2=92\frac{dy}{dx} = \frac{-8 - 1}{4 - 6} = \frac{-9}{-2} = \frac{9}{2}. Choice B incorrectly keeps the negative sign. Choice C uses −2x+y-2x + y instead of −2x−y-2x - y. Choice D makes both errors.

Question 14

If sin⁡(xy)+x2y=4\sin(xy) + x^2y = 4, then dydx\frac{dy}{dx} equals:

  1. −ycos⁡(xy)−2xyxcos⁡(xy)+x2\frac{-y\cos(xy) - 2xy}{x\cos(xy) + x^2} (correct answer)
  2. −cos⁡(xy)−2xycos⁡(xy)+x2\frac{-\cos(xy) - 2xy}{\cos(xy) + x^2}
  3. −ycos⁡(xy)−2xyxcos⁡(xy)+x2+sin⁡(xy)\frac{-y\cos(xy) - 2xy}{x\cos(xy) + x^2 + \sin(xy)}
  4. −xcos⁡(xy)−2xyycos⁡(xy)+x2\frac{-x\cos(xy) - 2xy}{y\cos(xy) + x^2}
Explanation: Differentiating both sides with respect to xx: ddx[sin⁡(xy)]+ddx[x2y]=0\frac{d}{dx}[\sin(xy)] + \frac{d}{dx}[x^2y] = 0. For the first term: cos⁡(xy)⋅ddx[xy]=cos⁡(xy)(y+xdydx)\cos(xy) \cdot \frac{d}{dx}[xy] = \cos(xy)(y + x\frac{dy}{dx}). For the second term: 2xy+x2dydx2xy + x^2\frac{dy}{dx}. So: cos⁡(xy)(y+xdydx)+2xy+x2dydx=0\cos(xy)(y + x\frac{dy}{dx}) + 2xy + x^2\frac{dy}{dx} = 0. Expanding: ycos⁡(xy)+xcos⁡(xy)dydx+2xy+x2dydx=0y\cos(xy) + x\cos(xy)\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 0. Collecting dydx\frac{dy}{dx} terms: (xcos⁡(xy)+x2)dydx=−ycos⁡(xy)−2xy(x\cos(xy) + x^2)\frac{dy}{dx} = -y\cos(xy) - 2xy. Choice B omits the yy coefficient in cos⁡(xy)\cos(xy) terms. Choice C incorrectly adds sin⁡(xy)\sin(xy) to the denominator. Choice D switches xx and yy coefficients.