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MathidiaMathematics Academy
Advanced differentiation
Lesson 29 of 75
Challenge

Implicit differentiation

Objective. Differentiate relations not solved for y.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Implicit differentiation finds dy/dx when y is not isolated as a function of x alone, such as in x² + y² = 25. Differentiate every term with respect to x, treating y as a function of x and multiplying each y-term's derivative by dy/dx via the chain rule, then rearrange to collect dy/dx. For x² + y² = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. Always remember the extra dy/dx factor whenever you differentiate a term containing y.

Worked example

Find dy/dx by implicit differentiation given x² + y³ = 10.

10
Illustrated card showing 10
  1. 1Differentiate both sides with respect to x: 2x + 3y²(dy/dx) = 0
  2. 2Isolate the dy/dx term: 3y²(dy/dx) = −2x
  3. 3Divide both sides by 3y²
  4. 4State dy/dx = −2x/(3y²)

Answer: dy/dx = −2x/(3y²)

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Common mistake. A common mistake is differentiating y² as 2y and forgetting the chain-rule factor dy/dx, writing 2x + 2y = 0 instead of 2x + 2y(dy/dx) = 0. Always attach dy/dx to every differentiated y-term.

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