Implicit Differentiation Explained - Product Rule, Quotient & Chain Rule - Calculus

Implicit Differentiation

Introduction to Implicit Differentiation

• Implicit differentiation is a technique used to differentiate equations that relate variables implicitly, rather than explicitly expressing one variable in terms of the other.

Example 1: Differentiate (x^3 + y^3 = 8)

• Step 1: Differentiate both sides with respect to (x).

  • (d/dx[x^3] = 3x^2)

  • (d/dx[y^3] = 3y^2 \cdot \frac{dy}{dx}) (Chain Rule)

  • (d/dx[8] = 0)

• Step 2: Set up the equation: [ 3y^2 \cdot \frac{dy}{dx} = -3x^2 ]

• Step 3: Isolate (\frac{dy}{dx}):[ \frac{dy}{dx} = -\frac{x^2}{y^2} ]

Example 2: Differentiate (x^2 + 2xy + y^2 = 5)

• Step 1: Differentiate both sides.

  • (d/dx[x^2] = 2x)

  • Using Product Rule on (2xy):

    • Differentiate (2x): gives (2) while keeping (y) the same.

    • Differentiate (y): gives (1 \cdot \frac{dy}{dx}) while keeping (2x) the same.

  • (d/dx[y^2] = 2y \cdot \frac{dy}{dx})

• Step 2: Combine the derivatives: [ 2x + 2y \cdot \frac{dy}{dx} + 2y \cdot \frac{dy}{dx} = 0 ]

• Step 3: Combine like terms and isolate (\frac{dy}{dx}): [ (2x + 2y) \cdot \frac{dy}{dx}= -2x ]

• Step 4: Divide to isolate (\frac{dy}{dx}):[ \frac{dy}{dx} = -1 ]

Example 3: Differentiate (5xy - y^3 = 8)

• Step 1: Differentiate both sides.

  • For (5xy), use the Product Rule.

    • (5 \cdot y + 5x \cdot \frac{dy}{dx})

  • (d/dx[y^3] = 3y^2 \cdot \frac{dy}{dx})

• Step 2: Set up the equation: [ 5y + 5x\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0 ]

• Step 3: Isolate (\frac{dy}{dx}): [ \frac{dy}{dx}(5x - 3y^2) = -5y ]

• Step 4: Divide to find (\frac{dy}{dx}): [ \frac{dy}{dx} = -\frac{5y}{5x - 3y^2} ]

Example 4: Differentiate (\tan(xy) = 7)

• Step 1: Differentiate both sides:

  • (d/dx[tan(xy)] = sec^2(xy) \cdot \frac{d}{dx}[xy])

• Step 2: Apply Product Rule on (xy):

  • (y \cdot 1 + x \cdot \frac{dy}{dx})

• Step 3: Set the equation: [ y \sec^2(xy) + x\frac{dy}{dx}\sec^2(xy) = 0 ]

• Step 4: Solve for (\frac{dy}{dx}):

  • Isolate and rearrange to find that (\frac{dy}{dx} = -\frac{y}{x})

Example 5: Differentiate (36 = \sqrt{x^2 + y^2})

• Step 1: Square both sides for simplicity: [ 36^2 = x^2 + y^2 ]

• Step 2: Differentiate both sides:

  • Derivative of (x^2) is (2x) and (d/dx[y^2] = 2y \cdot \frac{dy}{dx})

• Step 3: Set the equation and solve for (\frac{dy}{dx}):

• This leads to a direct and simplified working method for implicit differentiation.