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Implicit Differentiation - Business Calculus

  1. 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).
    Answer: 1/21/2 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}.
  2. 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).
    Answer: x=2x = 2 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.
  3. 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).
    Answer: 2/32/3 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}.
  4. Find the derivative dydx\frac{dy}{dx} for the implicit relation x2y=1+y2xx^2y = 1 + y^2x.
    Answer: y2−2xyx2−2xy\frac{y^2 - 2xy}{x^2 - 2xy} 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}.
  5. 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:
    Answer: dydx=−1\frac{dy}{dx} = -1 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.
  6. 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}).
    Answer: −23-2\sqrt{3} 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).
  7. At which yy-coordinates does the curve defined by x2−xy+y2=3x^2 - xy + y^2 = 3 have a vertical tangent line?
    Answer: y=1y = 1 and y=−1y = -1 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.
  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.
    Answer: −1/3-1/3 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}.
  9. 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?
    Answer: x=4x = 4 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.
  10. 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).
    Answer: 4/54/5 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}.
  11. 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.
    Answer: dydx=92\frac{dy}{dx} = \frac{9}{2} 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.
  12. If sin⁡(xy)+x2y=4\sin(xy) + x^2y = 4, then dydx\frac{dy}{dx} equals:
    Answer: −ycos⁡(xy)−2xyxcos⁡(xy)+x2\frac{-y\cos(xy) - 2xy}{x\cos(xy) + x^2} 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.
  13. 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?
    Answer: -3 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.
  14. 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).
    Answer: 55 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}.

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