AP Calculus AB · Lesson 2

Implicit differentiation

Know Know it

Some curves, such as x2+xy+y2=7x^2 + xy + y^2 = 7, are not written as y=f(x)y = f(x). Implicit differentiation finds dydx\frac{dy}{dx} anyway. Differentiate both sides with respect to xx, treating yy as a function of xx.

Each term with yy gets a factor of dydx\frac{dy}{dx} from the chain rule: ddxy2=2ydydx\frac{d}{dx}y^2 = 2y\frac{dy}{dx}. A term with both xx and yy needs the product rule: ddx(xy)=y+xdydx\frac{d}{dx}(xy) = y + x\frac{dy}{dx}.

Then collect the dydx\frac{dy}{dx} terms on one side, factor out dydx\frac{dy}{dx}, and divide. The answer usually has both xx and yy in it, so you need both coordinates of a point to find a slope there.

Worked example Easy

For the curve x2+y2=34x^2 + y^2 = 34, what is dydx\frac{dy}{dx}?

  1. −yx-\frac{y}{x}
  2. xy\frac{x}{y}
  3. −xy-\frac{x}{y}Answer
  4. −2x-2x

How to solve it

  1. Differentiate both sides: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.
  2. Move 2x2x over: 2ydydx=−2x2y\frac{dy}{dx} = -2x.
  3. Divide: dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.

Why each choice is right or wrong

  • A. Incorrect. −yx-\frac{y}{x} flips the fraction; solving 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0 puts yy in the denominator.
  • B. Incorrect. xy\frac{x}{y} loses the minus sign that comes from moving 2x2x to the other side.
  • C. Correct. Differentiating gives 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−2x2y=−xy\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}.
  • D. Incorrect. −2x-2x comes from 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0 with the factor 2y2y dropped, as if ddxy2\frac{d}{dx}y^2 were just dydx\frac{dy}{dx}; it is 2ydydx2y\frac{dy}{dx}, so you must divide by 2y2y.

Worked example Medium

What is the slope of the curve x2+xy+y2=7x^2 + xy + y^2 = 7 at the point (1,2)(1, 2)?

  1. −54-\frac{5}{4}
  2. −45-\frac{4}{5}Answer
  3. −25-\frac{2}{5}
  4. 45\frac{4}{5}

How to solve it

  1. Differentiate: 2x+(y+xdydx)+2ydydx=02x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0.
  2. Collect: (x+2y)dydx=−(2x+y)(x + 2y)\frac{dy}{dx} = -(2x + y), so dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x + y}{x + 2y}.
  3. At (1,2)(1, 2): −2+21+4=−45-\frac{2 + 2}{1 + 4} = -\frac{4}{5}.

Why each choice is right or wrong

  • A. Incorrect. −54-\frac{5}{4} turns the fraction upside down: the dydx\frac{dy}{dx} terms x+2yx + 2y belong in the denominator.
  • B. Correct. Differentiating gives 2x+y+xdydx+2ydydx=02x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0, so dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x + y}{x + 2y}. At (1,2)(1, 2) this is −45-\frac{4}{5}.
  • C. Incorrect. −25-\frac{2}{5} treats xyxy as if its derivative were only xdydxx\frac{dy}{dx}, dropping the yy from the product rule.
  • D. Incorrect. 45\frac{4}{5} loses the minus sign when the xx-only terms are moved to the other side.

Explain Explain it

Task verbs: explain, determine

Write dydx=ND\frac{dy}{dx} = \frac{N}{D}. The tangent is horizontal where N=0N = 0 and D≠0D \ne 0, and vertical where D=0D = 0 and N≠0N \ne 0. Any candidate point must also lie on the curve, so substitute back into the original equation.

For a second derivative implicitly, differentiate dydx\frac{dy}{dx} with respect to xx, using the quotient rule if needed. Every yy brings a new dydx\frac{dy}{dx}. Then replace each dydx\frac{dy}{dx} with its formula, and simplify using the original equation if it helps.

Trap Every y term gets dy/dx

The most common slip is writing ddxy3=3y2\frac{d}{dx}y^3 = 3y^2. Because yy changes with xx, the chain rule gives 3y2dydx3y^2\frac{dy}{dx}. A constant, such as the 7 in x2+xy+y2=7x^2 + xy + y^2 = 7, has derivative 0.

Worked example Hard

For the curve x2+y2=34x^2 + y^2 = 34, what is d2ydx2\frac{d^2y}{dx^2} at the point (3,5)(3, 5)?

  1. −34125-\frac{34}{125}Answer
  2. −35-\frac{3}{5}
  3. −15-\frac{1}{5}
  4. −16125-\frac{16}{125}

How to solve it

  1. First derivative: dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.
  2. Quotient rule: d2ydx2=−y⋅1−xdydxy2=−y+x2yy2=−x2+y2y3\frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x\frac{dy}{dx}}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}.
  3. On the curve x2+y2=34x^2 + y^2 = 34, so d2ydx2=−34y3=−34125\frac{d^2y}{dx^2} = -\frac{34}{y^3} = -\frac{34}{125} at y=5y = 5.

Why each choice is right or wrong

  • A. Correct. y′=−xyy' = -\frac{x}{y}, so y′′=−y−xy′y2=−y2+x2y3=−34y3y'' = -\frac{y - xy'}{y^2} = -\frac{y^2 + x^2}{y^3} = -\frac{34}{y^3}, which is −34125-\frac{34}{125} at (3,5)(3, 5).
  • B. Incorrect. −35-\frac{3}{5} is dydx\frac{dy}{dx} at (3,5)(3, 5), the first derivative, not the second.
  • C. Incorrect. −15-\frac{1}{5} differentiates −xy-\frac{x}{y} as if yy were a constant, giving −1y-\frac{1}{y}; yy changes with xx.
  • D. Incorrect. −16125-\frac{16}{125} takes xdydxx\frac{dy}{dx} as +x2y+\frac{x^2}{y}, which gives −y2−x2y3-\frac{y^2 - x^2}{y^3}; in fact xdydx=−x2yx\frac{dy}{dx} = -\frac{x^2}{y}, so the top is y2+x2y^2 + x^2.

Determine Your turn

For the curve y2−4x2=9y^2 - 4x^2 = 9, find dydx\frac{dy}{dx}. Then find each point on the curve where the tangent line is horizontal.

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A full answer does this

  • Finds dydx=4xy\frac{dy}{dx} = \frac{4x}{y}
  • Finds both points with a horizontal tangent, (0,3)(0, 3) and (0,−3)(0, -3)

Model answer

Differentiating gives 2ydydx−8x=02y\frac{dy}{dx} - 8x = 0, so dydx=8x2y=4xy\frac{dy}{dx} = \frac{8x}{2y} = \frac{4x}{y}. The tangent is horizontal where the top is 0 and the bottom is not: x=0x = 0. Then y2=9y^2 = 9, so y=3y = 3 or y=−3y = -3, and neither makes the bottom 0. The points are (0,3)(0, 3) and (0,−3)(0, -3).

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