Lecture Notes on Implicit Differentiation✅

Chain Rule Recall

  • When taking the derivative of a composite function f(g(x)), the formula is: \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x). This is likened to peeling a mandarin.

Examples of Chain Rule

  • Example 1: Derivative of g(x)^2
    • Think of this as f(g(x)) where f(x) = x^2.
    • The derivative is then 2 \cdot g(x) \cdot g'(x).
  • Example 2: Derivative of \ln(g(x))
    • Think of this as f(g(x)) where f(x) = \ln(x).
    • The derivative is \frac{1}{g(x)} \cdot g'(x).
    • This specific example will be beneficial in the upcoming discussion of logarithmic differentiation.

Logarithmic Differentiation

  • Consider a function y = f(x)^\alpha \cdot g(x)^\beta, where \alpha and \beta are constants.
  • Take the natural logarithm of both sides: \ln(y) = \ln(f(x)^\alpha \cdot g(x)^\beta)
  • Apply logarithm properties:
    • \ln(x \cdot y) = \ln(x) + \ln(y)
    • \ln(x^\alpha) = \alpha \cdot \ln(x)
  • This simplifies to: \ln(y) = \alpha \ln(f(x)) + \beta \ln(g(x))
  • Now, differentiate both sides with respect to x, using the chain rule.

Applying the Chain Rule

  • Starting with: \ln(y) = \alpha \ln(f(x)) + \beta \ln(g(x))
  • Differentiate with respect to x:
    • \frac{1}{y} \cdot \frac{dy}{dx} = \alpha \cdot \frac{1}{f(x)} \cdot f'(x) + \beta \cdot \frac{1}{g(x)} \cdot g'(x)
  • Solve for \frac{dy}{dx}:
    • \frac{dy}{dx} = y \cdot [\alpha \cdot \frac{f'(x)}{f(x)} + \beta \cdot \frac{g'(x)}{g(x)}]
    • Substitute y: \frac{dy}{dx} = f(x)^\alpha \cdot g(x)^\beta \cdot [\alpha \cdot \frac{f'(x)}{f(x)} + \beta \cdot \frac{g'(x)}{g(x)}]

Example 1

  • Given: y = (x^2 + 1) \cdot \sqrt{x^2 + 2} = (x^2 + 1) \cdot (x^2 + 2)^{\frac{1}{2}}
  • Take the natural logarithm of both sides: \ln(y) = \ln(x^2 + 1) + \frac{1}{2}\ln(x^2 + 2)
  • Differentiate implicitly using the chain rule:
    • \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot 2x + \frac{1}{2} \cdot \frac{1}{x^2 + 2} \cdot 2x
  • Solve for \frac{dy}{dx}:
    • \frac{dy}{dx} = (x^2 + 1) \cdot \sqrt{x^2 + 2} \cdot [\frac{2x}{x^2 + 1} + \frac{x}{x^2 + 2}]

Example 2

  • Given: y = x^x
  • Warning: The derivative is not x \cdot x^{x-1}. This is a common mistake because both the base and exponent are variables.
  • Take the natural logarithm of both sides: \ln(y) = \ln(x^x) = x \cdot \ln(x)
  • Differentiate both sides with respect to x, using the chain and product rules:
    • \frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1
  • Solve for \frac{dy}{dx}:
    • \frac{dy}{dx} = y \cdot (\ln(x) + 1) = x^x \cdot (\ln(x) + 1)

Comparison

  • For y = x^x, logarithmic differentiation is essential.
  • There is no direct formula to find the derivative when both the base and exponent are variables.

Implicit Differentiation

  • Consider a curve defined by an equation involving both x and y, such as x^2 + y^2 = a^2 (a circle).
  • A circle is not a function, but we can solve for y to get two functions: y = \sqrt{a^2 - x^2} and y = -\sqrt{a^2 - x^2}.
  • Sometimes, it's not possible to explicitly solve for y in terms of x. In such cases, we use implicit differentiation.

Process

  • Differentiate both sides of the equation with respect to x.
  • Treat y as an implicit function of x, i.e., y = y(x).
  • Use the chain rule when differentiating terms involving y.
  • Solve for \frac{dy}{dx}.

Example: Circle

  • Given: x^2 + y^2 = a^2
  • Differentiate both sides with respect to x:
    • \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(a^2)
    • 2x + 2y \cdot \frac{dy}{dx} = 0
  • Solve for \frac{dy}{dx}:
    • \frac{dy}{dx} = -\frac{x}{y}

Chain Rule Explanation

  • The derivative of y^2 with respect to x is found using the chain rule:
    • Let h(y) = y^2, then \frac{d}{dx}(h(y)) = h'(y) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}.

Example: Curve

  • Given: x^2 + 3xy + 2y^3 = 6
  • Find \frac{dy}{dx}.
  • Differentiate both sides with respect to x:
    • \frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) + \frac{d}{dx}(2y^3) = \frac{d}{dx}(6)
    • 2x + 3(x\cdot \frac{dy}{dx} + y \cdot 1) + 6y^2 \cdot \frac{dy}{dx} = 0
  • Apply the product rule to the 3xy term.
  • Solve for \frac{dy}{dx}:
    • 2x + 3y + 3x\frac{dy}{dx} + 6y^2\frac{dy}{dx} = 0
    • (3x + 6y^2)\frac{dy}{dx} = -2x - 3y
    • \frac{dy}{dx} = \frac{-2x - 3y}{3x + 6y^2}

Tangent Line

  • To find the tangent line at a point (1, 1) on the curve:
    • Evaluate \frac{dy}{dx} at (1, 1): \frac{dy}{dx}|_{(1,1)} = \frac{-2(1) - 3(1)}{3(1) + 6(1)^2} = -\frac{5}{9}
    • The equation of the tangent line is: y - 1 = -\frac{5}{9}(x - 1).

Higher Derivatives

  • If a function f(x) is differentiable and its derivative is also differentiable, we can find the second derivative.
  • Notation:
    • f''(x)
    • \frac{d^2f}{dx^2}
  • We can continue this process to find higher-order derivatives if they exist.

Example

  • Given: f(x) = x^4
  • First derivative: f'(x) = 4x^3
  • Second derivative: f''(x) = 12x^2
  • Third derivative: f'''(x) = 24x

Meaning of the Second Derivative

  • The second derivative provides information about the concavity of a function's graph.
  • If f''(x) > 0, the function is concave upwards in that interval.
  • If f''(x) < 0, the function is concave downwards in that interval.

Inflection Points

  • An inflection point occurs where the concavity of a function changes.
  • This happens when f''(x) = 0 or is undefined, and the concavity changes sign around that point.

Examples

  • y = x^2: y'' = 2 (concave upwards everywhere).
  • y = -x^2: y'' = -2 (concave downwards everywhere).
  • y = x^3: y'' = 6x
    • If x < 0, y'' < 0 (concave downwards).
    • If x > 0, y'' > 0 (concave upwards).
    • At x = 0, there is an inflection point.

Graph of y = x^3

  • The graph visually shows the change in concavity at the inflection point (0, 0).
  • The function is concave downwards for x < 0 and concave upwards for x > 0.

Conclusion

  • Summary of logarithmic differentiation, implicit differentiation and higher derivatives.