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Differentiation

2.5 Implicit Differentiation

Focus on implicit first derivatives, second implicit derivatives, tangent and normal lines.

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Ideas in this lesson

01Implicit first derivativesWorked steps · five questions02Second implicit derivativesWorked steps · five questions03Tangent and normal linesWorked steps · five questions
Idea 1 of 3

Implicit first derivatives

Treat yy as a function of xx. Every differentiated yy term needs the appropriate y′y' factor.

Worked example 1

On x2+y2=4x^{2} + y^{2} = 4, find dydx\dfrac{dy}{dx} at (0,2)(0, 2).
  1. 2x+22x + 2 yy ′=0' = 0
  2. y′=−xyy' = -\dfrac{x}{y}
  3. y′(0,2)=−02=0y'(0, 2) = -\dfrac{0}{2} = 0

Answer: 00

Worked example 2

On x2+y2=25x^{2} + y^{2} = 25, find dydx\dfrac{dy}{dx} at (3,4)(3, 4).
  1. 2x+22x + 2 yy ′=0' = 0
  2. y′=−xyy' = -\dfrac{x}{y}
  3. y′(3,4)=−34=−34≈−0.75y'(3, 4) = -\dfrac{3}{4} = \dfrac{-3}{4} \approx -0.75

Answer: −34≈−0.75\dfrac{-3}{4} \approx -0.75

Idea 2 of 3

Second implicit derivatives

Differentiate the first derivative relation again, using the product or quotient rule as needed.

Worked example 1

For x2+y2=4x^{2} + y^{2} = 4, find y′y' at (0,2)(0, 2).
  1. y′=−xyy' = -\dfrac{x}{y}
  2. y′′=−y2+x2y3y'' = -\dfrac{y^{2} + x^{2}}{y^{3}}
  3. y′(0,2)=0y'(0, 2) = 0

Answer: 00

Worked example 2

For x2+y2=9x^{2} + y^{2} = 9, find y′′y'' at (0,3)(0, 3).
  1. y′=−xyy' = -\dfrac{x}{y}
  2. y′′=−y2+x2y3y'' = -\dfrac{y^{2} + x^{2}}{y^{3}}
  3. y′′(0,3)=−927=−13≈−0.33y''(0, 3) = -\dfrac{9}{27} = \dfrac{-1}{3} \approx -0.33

Answer: −13≈−0.33\dfrac{-1}{3} \approx -0.33

Idea 3 of 3

Tangent and normal lines

Evaluate an implicit derivative only at a point on the curve. A horizontal tangent has a vertical normal.

Worked example 1

On x2+y2=4x^{2} + y^{2} = 4, find the tangent slope at (0,2)(0, 2).
  1. 2x+22x + 2 yy ′=0  ⟹  y′=−xy' = 0 \implies y' = -\dfrac{x}{y}
  2. y′=−02=0y' = -\dfrac{0}{2} = 0

Answer: 00

Worked example 2

On x2+y2=25x^{2} + y^{2} = 25, find the tangent slope at (3,4)(3, 4).
  1. 2x+22x + 2 yy ′=0  ⟹  y′=−xy' = 0 \implies y' = -\dfrac{x}{y}
  2. y′=−34=−34≈−0.75y' = -\dfrac{3}{4} = \dfrac{-3}{4} \approx -0.75

Answer: −34≈−0.75\dfrac{-3}{4} \approx -0.75

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §2.5,p2.5, p. 140140. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.