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.