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Implicit differentiation: differentiate both sides, then solve for dy/dx

Ten implicit derivatives: the slope and tangent line of a circle and an ellipse, a curve with an xy term, an exponential and a log, a trig relation, the folium of Descartes, the horizontal and vertical tangents of a tilted ellipse, a second derivative, the derivative of arcsin, and logarithmic differentiation of a product and quotient, with worked solutions and the mistakes of dropping dy/dx, skipping the product rule, differentiating a constant, losing a y' term and treating y as a constant.

Before you start

Some curves are given by an equation in xx and yy that is hard or impossible to solve for yy, such as x3+y3=6xyx^3 + y^3 = 6xy. Implicit differentiation finds the slope anyway: treat yy as a function of xx, differentiate both sides of the equation with respect to xx, and solve the result for y′y'. It is the same chain rule that drives related rates, and the idea behind differentiating through a quantity defined by an equation, such as the optimum of an inner problem, instead of by a formula. The work goes wrong in a few places: a y′y' dropped from the chain rule, a product rule skipped on xyxy, a constant side differentiated as if it were a variable, a y′y' term left on the wrong side, and yy treated as a constant in a second derivative. These ten problems cover each of them.

  • yy is a function of xx, and y′y' means dydx\tfrac{dy}{dx}. Every term containing yy is a composition, so its derivative carries a factor y′y': ddx y2=2y y′\tfrac{d}{dx}\,y^2 = 2y\,y', ddx sin⁡y=cos⁡y y′\tfrac{d}{dx}\,\sin y = \cos y\,y', ddx ey=ey y′\tfrac{d}{dx}\,e^{y} = e^{y}\,y', ddxln⁡y=y′y\tfrac{d}{dx}\ln y = \tfrac{y'}{y}.
  • A term with both variables needs the product rule as well: ddx(xy)=y+x y′\tfrac{d}{dx}(xy) = y + x\,y'. A constant differentiates to 0.
  • The method: (1) differentiate both sides of the equation with respect to xx; (2) move every term containing y′y' to one side and everything else to the other; (3) factor out y′y' and divide. Substitute the point's coordinates at the end, or after step 1 if only a number is wanted.
  • The slope usually depends on both xx and yy. The tangent line at (x0,y0)(x_0, y_0) is y−y0=m(x−x0)y - y_0 = m(x - x_0) with m=y′m = y' there. A tangent is horizontal where y′=0y' = 0 and vertical where the denominator of y′y' is 00 and the numerator is not.
  • Angles are in radians.

Builds on: Differentiation rules: chain, product, quotient

Problems

  1. ·

    Find y′y' for the circle x2+y2=25x^2 + y^2 = 25, and the tangent line at (3,4)(3, 4).

  2. ·

    Find y′y' for the ellipse x2+4y2=8x^2 + 4y^2 = 8, and its value at (2,1)(2, 1) and at (2,−1)(2, -1).

  3. ··

    Find y′y' for x2+3xy+y2=11x^2 + 3xy + y^2 = 11 at (1,2)(1, 2).

  4. ··

    Find y′y' for y ex+ln⁡y=1y\,e^{x} + \ln y = 1 at the point where x=0x = 0.

  5. ··

    Find y′y' for xsin⁡y+y=πx\sin y + y = \pi at (π2,π2)\big(\tfrac\pi2, \tfrac\pi2\big).

  6. ···

    The folium of Descartes is x3+y3=6xyx^3 + y^3 = 6xy. Find the slope and the tangent line at (3,3)(3, 3).

  7. ···

    Find the points on x2−xy+y2=3x^2 - xy + y^2 = 3 where the tangent is horizontal, and where it is vertical.

  8. ···

    For x2+y2=25x^2 + y^2 = 25, find y′′y'' in terms of yy alone, and its value at (3,4)(3, 4).

  9. ··

    Use sin⁡y=x\sin y = x to find the derivative of y=arcsin⁡xy = \arcsin x, and its value at x=12x = \tfrac12.

  10. ···

    Use logarithmic differentiation to find y′y' for y=(x+1)2(x+2)3(x+3)4y = \dfrac{(x + 1)^2(x + 2)^3}{(x + 3)^4}, and its value at x=0x = 0.

Answers

  1. y′=−xyy' = -\dfrac xy; tangent y=−34x+254y = -\tfrac34x + \tfrac{25}4
  2. y′=−x4yy' = -\dfrac{x}{4y}; −12-\tfrac12 at (2,1)(2, 1) and 12\tfrac12 at (2,−1)(2, -1)
  3. y′=−2+63+4=−87y' = -\dfrac{2 + 6}{3 + 4} = -\tfrac87
  4. y′=−12y' = -\tfrac12
  5. y′=−10+1=−1y' = -\dfrac{1}{0 + 1} = -1
  6. Slope −1-1; tangent y=−x+6y = -x + 6
  7. Horizontal at (1,2)(1, 2) and (−1,−2)(-1, -2); vertical at (2,1)(2, 1) and (−2,−1)(-2, -1)
  8. y′′=−25y3y'' = -\dfrac{25}{y^3}; at (3,4)(3, 4), −2564≈−0.391-\tfrac{25}{64} \approx -0.391
  9. At x=12x = \tfrac12: 13/4=23≈1.155\dfrac{1}{\sqrt{3/4}} = \dfrac{2}{\sqrt3} \approx 1.155
  10. y′=881⋅136=52243≈0.214y' = \tfrac{8}{81}\cdot\tfrac{13}{6} = \tfrac{52}{243} \approx 0.214

Worked solutions

Problem 1

Find y′y' for the circle x2+y2=25x^2 + y^2 = 25, and the tangent line at (3,4)(3, 4).

  1. 2x+2y y′=02x + 2y\,y' = 0Differentiate both sides; ddx y2=2y y′\tfrac{d}{dx}\,y^2 = 2y\,y' by the chain rule, and 2525 gives 00.
  2. y′=−xyy' = -\dfrac{x}{y}Move 2x2x across and divide by 2y2y.
  3. At (3,4)(3, 4): y′=−34y' = -\tfrac34Substitute.
  4. y−4=−34(x−3)y - 4 = -\tfrac34(x - 3)Point-slope form.
  5. y′=−xyy' = -\dfrac xy; tangent y=−34x+254y = -\tfrac34x + \tfrac{25}4The tangent is perpendicular to the radius, whose slope is 43\tfrac43.

Problem 2

Find y′y' for the ellipse x2+4y2=8x^2 + 4y^2 = 8, and its value at (2,1)(2, 1) and at (2,−1)(2, -1).

  1. 2x+8y y′=02x + 8y\,y' = 0Differentiate both sides.
  2. y′=−x4yy' = -\dfrac{x}{4y}Solve for y′y'.
  3. At (2,1)(2, 1): y′=−24=−12y' = -\tfrac{2}{4} = -\tfrac12. At (2,−1)(2, -1): y′=12y' = \tfrac12.Same xx, two points on the curve.
  4. y′=−x4yy' = -\dfrac{x}{4y}; −12-\tfrac12 at (2,1)(2, 1) and 12\tfrac12 at (2,−1)(2, -1)The slope needs both coordinates, because the ellipse is not the graph of one function of xx.

Problem 3

Find y′y' for x2+3xy+y2=11x^2 + 3xy + y^2 = 11 at (1,2)(1, 2).

  1. 1+6+4=111 + 6 + 4 = 11The point is on the curve.
  2. 2x+(3y+3x y′)+2y y′=02x + (3y + 3x\,y') + 2y\,y' = 0The product rule on 3xy3xy gives 3y+3x y′3y + 3x\,y'.
  3. y′(3x+2y)=−(2x+3y)y'(3x + 2y) = -(2x + 3y)Collect the y′y' terms on one side.
  4. y′=−2x+3y3x+2yy' = -\dfrac{2x + 3y}{3x + 2y}Divide.
  5. y′=−2+63+4=−87y' = -\dfrac{2 + 6}{3 + 4} = -\tfrac87Substitute x=1x = 1, y=2y = 2.

Problem 4

Find y′y' for y ex+ln⁡y=1y\,e^{x} + \ln y = 1 at the point where x=0x = 0.

  1. At x=0x = 0: y+ln⁡y=1y + \ln y = 1, so y=1y = 1.1+ln⁡1=11 + \ln 1 = 1, and y+ln⁡yy + \ln y is increasing, so 11 is the only solution.
  2. y′ ex+y ex+y′y=0y'\,e^{x} + y\,e^{x} + \dfrac{y'}{y} = 0Product rule on y exy\,e^x; chain rule on ln⁡y\ln y.
  3. y′(ex+1y)=−y exy'\Big(e^{x} + \dfrac1y\Big) = -y\,e^{x}Collect the y′y' terms.
  4. y′(1+1)=−1y'(1 + 1) = -1Substitute x=0x = 0, y=1y = 1.
  5. y′=−12y' = -\tfrac12Solving for yy first is not possible in elementary functions, and it was never needed.

Problem 5

Find y′y' for xsin⁡y+y=πx\sin y + y = \pi at (π2,π2)\big(\tfrac\pi2, \tfrac\pi2\big).

  1. π2⋅1+π2=π\tfrac\pi2\cdot1 + \tfrac\pi2 = \piThe point is on the curve.
  2. sin⁡y+xcos⁡y  y′+y′=0\sin y + x\cos y\;y' + y' = 0Product rule on xsin⁡yx\sin y, with ddxsin⁡y=cos⁡y y′\tfrac{d}{dx}\sin y = \cos y\,y'; the constant π\pi gives 00.
  3. y′=−sin⁡yxcos⁡y+1y' = -\dfrac{\sin y}{x\cos y + 1}Collect and divide.
  4. y′=−10+1=−1y' = -\dfrac{1}{0 + 1} = -1sin⁡π2=1\sin\tfrac\pi2 = 1 and cos⁡π2=0\cos\tfrac\pi2 = 0.

Problem 6

The folium of Descartes is x3+y3=6xyx^3 + y^3 = 6xy. Find the slope and the tangent line at (3,3)(3, 3).

  1. 27+27=54=6⋅927 + 27 = 54 = 6\cdot9The point is on the curve.
  2. 3x2+3y2 y′=6y+6x y′3x^2 + 3y^2\,y' = 6y + 6x\,y'Chain rule on y3y^3; product rule on the right.
  3. y′(3y2−6x)=6y−3x2y'(3y^2 - 6x) = 6y - 3x^2Both sides have a y′y' term: collect them on the left.
  4. y′=2y−x2y2−2xy' = \dfrac{2y - x^2}{y^2 - 2x}Divide, and cancel the common factor 3.
  5. y′=6−99−6=−1y' = \dfrac{6 - 9}{9 - 6} = -1Substitute.
  6. Slope −1-1; tangent y=−x+6y = -x + 6y−3=−(x−3)y - 3 = -(x - 3). The curve is symmetric in xx and yy, and (3,3)(3, 3) is on the line of symmetry, so a slope of −1-1 is expected.

Problem 7

Find the points on x2−xy+y2=3x^2 - xy + y^2 = 3 where the tangent is horizontal, and where it is vertical.

  1. 2x−(y+x y′)+2y y′=02x - (y + x\,y') + 2y\,y' = 0Product rule on xyxy, with its minus sign.
  2. y′=y−2x2y−xy' = \dfrac{y - 2x}{2y - x}y′(2y−x)=y−2xy'(2y - x) = y - 2x.
  3. Horizontal: y=2xy = 2x, so x2−2x2+4x2=3x2=3x^2 - 2x^2 + 4x^2 = 3x^2 = 3 and x=±1x = \pm1.Numerator 00; substitute y=2xy = 2x into the curve.
  4. Vertical: x=2yx = 2y, so 4y2−2y2+y2=3y2=34y^2 - 2y^2 + y^2 = 3y^2 = 3 and y=±1y = \pm1.Denominator 00; substitute x=2yx = 2y.
  5. Horizontal at (1,2)(1, 2) and (−1,−2)(-1, -2); vertical at (2,1)(2, 1) and (−2,−1)(-2, -1)At each point the other part of the fraction is not 00: at (1,2)(1, 2) the denominator is 33, at (2,1)(2, 1) the numerator is −3-3.

Problem 8

For x2+y2=25x^2 + y^2 = 25, find y′′y'' in terms of yy alone, and its value at (3,4)(3, 4).

  1. y′=−xyy' = -\dfrac xyProblem 1.
  2. y′′=−1⋅y−x y′y2y'' = -\dfrac{1\cdot y - x\,y'}{y^2}Quotient rule; yy is still a function of xx, so its derivative is y′y', not 00.
  3. y′′=−y+x2/yy2=−y2+x2y3y'' = -\dfrac{y + x^2/y}{y^2} = -\dfrac{y^2 + x^2}{y^3}Substitute y′=−xyy' = -\tfrac xy and multiply top and bottom by yy.
  4. y′′=−25y3y'' = -\dfrac{25}{y^3}Use the curve: x2+y2=25x^2 + y^2 = 25.
  5. y′′=−25y3y'' = -\dfrac{25}{y^3}; at (3,4)(3, 4), −2564≈−0.391-\tfrac{25}{64} \approx -0.391Negative on the upper half, where the circle is concave down.

Problem 9

Use sin⁡y=x\sin y = x to find the derivative of y=arcsin⁡xy = \arcsin x, and its value at x=12x = \tfrac12.

  1. cos⁡y  y′=1\cos y\;y' = 1Differentiate sin⁡y=x\sin y = x.
  2. y′=1cos⁡yy' = \dfrac{1}{\cos y}Divide.
  3. cos⁡y=1−sin⁡2y=1−x2\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}arcsin⁡\arcsin takes values in [−π2,π2][-\tfrac\pi2, \tfrac\pi2], where cos⁡y≥0\cos y \ge 0, so the positive root is the right one.
  4. ddxarcsin⁡x=11−x2\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1 - x^2}}For −1<x<1-1 < x < 1.
  5. At x=12x = \tfrac12: 13/4=23≈1.155\dfrac{1}{\sqrt{3/4}} = \dfrac{2}{\sqrt3} \approx 1.155The same steps on tan⁡y=x\tan y = x give 11+x2\tfrac{1}{1 + x^2} for arctan⁡\arctan.

Problem 10

Use logarithmic differentiation to find y′y' for y=(x+1)2(x+2)3(x+3)4y = \dfrac{(x + 1)^2(x + 2)^3}{(x + 3)^4}, and its value at x=0x = 0.

  1. ln⁡y=2ln⁡(x+1)+3ln⁡(x+2)−4ln⁡(x+3)\ln y = 2\ln(x + 1) + 3\ln(x + 2) - 4\ln(x + 3)Take logs (for x>−1x > -1, where every factor is positive): products become sums and powers become coefficients.
  2. y′y=2x+1+3x+2−4x+3\dfrac{y'}{y} = \dfrac{2}{x + 1} + \dfrac{3}{x + 2} - \dfrac{4}{x + 3}Differentiate; the left side is a chain rule on ln⁡y\ln y.
  3. y′=y(2x+1+3x+2−4x+3)y' = y\Big(\dfrac{2}{x + 1} + \dfrac{3}{x + 2} - \dfrac{4}{x + 3}\Big)Multiply by yy, then write yy out in full.
  4. At x=0x = 0: y=1⋅881=881y = \tfrac{1\cdot8}{81} = \tfrac{8}{81} and y′y=2+32−43=136\tfrac{y'}{y} = 2 + \tfrac32 - \tfrac43 = \tfrac{13}6.Substitute.
  5. y′=881⋅136=52243≈0.214y' = \tfrac{8}{81}\cdot\tfrac{13}{6} = \tfrac{52}{243} \approx 0.214The quotient rule gets the same answer through a much longer numerator.

Where this goes wrong

1. Dropping the y' on a y term

Differentiating y2y^2 gives 2y2y in every exercise where yy is the variable.

  1. x2+3xy+y2=11x^2 + 3xy + y^2 = 11 at (1,2)(1, 2)Right so far: Problem 3.
  2. “ddx y2=2y\tfrac{d}{dx}\,y^2 = 2y.”The habit that causes the mistake: differentiating y2y^2 as if yy were the variable of differentiation.
  3. 2x+3y+3x y′+2y=02x + 3y + 3x\,y' + 2y = 0, so y′=−2x+5y3x=−4y' = -\dfrac{2x + 5y}{3x} = -4yy depends on xx, so ddx y2=2y y′\tfrac{d}{dx}\,y^2 = 2y\,y'. With the factor, y′(3x+2y)=−(2x+3y)y'(3x + 2y) = -(2x + 3y) and y′=−87y' = -\tfrac87. Every term with a yy in it must produce a y′y'.

2. Skipping the product rule on xy

3xy3xy looks like a single term, so it gets a single derivative.

  1. x2+3xy+y2=11x^2 + 3xy + y^2 = 11 at (1,2)(1, 2)Right so far: Problem 3.
  2. “ddx(3xy)=3x y′\tfrac{d}{dx}(3xy) = 3x\,y'.”The habit that causes the mistake: treating 3x3x as a constant coefficient of yy.
  3. 2x+3x y′+2y y′=02x + 3x\,y' + 2y\,y' = 0, so y′=−2x3x+2y=−27y' = -\dfrac{2x}{3x + 2y} = -\tfrac27xx changes too, so 3xy3xy is a product: ddx(3xy)=3y+3x y′\tfrac{d}{dx}(3xy) = 3y + 3x\,y'. The missing 3y3y changes the numerator from 2x+3y=82x + 3y = 8 to 2x=22x = 2, and the answer is −87-\tfrac87, four times as steep.

3. Differentiating the constant side

The right side is π\pi, and π\pi looks like an angle, so it is treated like a variable.

  1. xsin⁡y+y=πx\sin y + y = \pi at (π2,π2)\big(\tfrac\pi2, \tfrac\pi2\big)Right so far: Problem 5.
  2. “ddx π=1\tfrac{d}{dx}\,\pi = 1.”The habit that causes the mistake: differentiating every symbol that is not a digit.
  3. sin⁡y+xcos⁡y  y′+y′=1\sin y + x\cos y\;y' + y' = 1, so 1+y′=11 + y' = 1 and y′=0y' = 0π\pi is a number, and the derivative of a number is 00. With 00 on the right, 1+y′=01 + y' = 0 and y′=−1y' = -1. A horizontal tangent here would be a coincidence; check the answer against the curve instead.

4. Leaving a y' term on the other side

After differentiating, the left side has a y′y' and is solved for it, the way an explicit derivative would be.

  1. 3x2+3y2 y′=6y+6x y′3x^2 + 3y^2\,y' = 6y + 6x\,y' at (3,3)(3, 3)Right so far: Problem 6, step 2.
  2. “y′=6y−3x23y2y' = \dfrac{6y - 3x^2}{3y^2}.”The habit that causes the mistake: isolating the y′y' on the left and treating the right side as known.
  3. y′=18−2727=−13y' = \dfrac{18 - 27}{27} = -\tfrac13The right side still contains 6x y′6x\,y', so this is not a solution for y′y': y′y' appears on both sides. Collect the terms first, y′(3y2−6x)=6y−3x2y'(3y^2 - 6x) = 6y - 3x^2, and the slope is −99=−1\tfrac{-9}{9} = -1.

5. Treating y as a constant in the second derivative

In −xy-\tfrac xy the yy sits in the denominator like a constant, so only the xx gets differentiated.

  1. y′=−xyy' = -\dfrac xy on x2+y2=25x^2 + y^2 = 25, at (3,4)(3, 4)Right so far: Problem 8, step 1.
  2. “y′′=−1yy'' = -\dfrac1y.”The habit that causes the mistake: forgetting that yy is a function of xx once the first derivative is done.
  3. y′′=−14y'' = -\tfrac14 at (3,4)(3, 4)The quotient rule needs the derivative of the denominator, which is y′=−34y' = -\tfrac34, not 00: y′′=−y−x y′y2=−4+9/416=−2564y'' = -\tfrac{y - x\,y'}{y^2} = -\tfrac{4 + 9/4}{16} = -\tfrac{25}{64}. Then substitute y′y' and use the curve to simplify, as in Problem 8.

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