Implicit Differentiation, Tangents, Logarithmic Differentiation, and Related Rates

Implicit Differentiation and Related Rates

Implicit Differentiation: Concept and Method

Implicit differentiation is a technique used when an equation defines y implicitly in terms of x, rather than as an explicit function. This method is essential for finding derivatives when y cannot be easily isolated.

• Implicit Equation: An equation where y is not solved explicitly as a function of x (e.g., $4x^2 + 24x + 9y^2 + 36y = 36$).

• Procedure:

  1. Apply $\frac{d}{dx}$ to both sides of the equation.

  2. When differentiating terms involving y, use the chain rule: multiply by $\frac{dy}{dx}$.

  3. Collect all $\frac{dy}{dx}$ terms on one side and factor.

  4. Solve algebraically for $\frac{dy}{dx}$.

• Chain Rule: If $y = y(x)$, then $\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$.

Example: Differentiate $4x^2 + 24x + 9y^2 + 36y = 36$ with respect to $x$:

• $\frac{d}{dx}(4x^2) = 8x$

• $\frac{d}{dx}(24x) = 24$

• $\frac{d}{dx}(9y^2) = 18y \cdot \frac{dy}{dx}$

• $\frac{d}{dx}(36y) = 36 \cdot \frac{dy}{dx}$

• $\frac{d}{dx}(36) = 0$

Combine and solve for $\frac{dy}{dx}$:

• $8x + 24 + 18y \cdot \frac{dy}{dx} + 36 \cdot \frac{dy}{dx} = 0$

• Group $\frac{dy}{dx}$ terms: $18y \cdot \frac{dy}{dx} + 36 \cdot \frac{dy}{dx} = -8x - 24$

• Factor: $(18y + 36) \cdot \frac{dy}{dx} = -8x - 24$

• Solve: $\frac{dy}{dx} = \frac{-8x - 24}{18y + 36}$

Finding Horizontal and Vertical Tangents

Implicit derivatives can be used to find points where the curve has horizontal or vertical tangents.

• Horizontal Tangent: Occurs where $\frac{dy}{dx} = 0$ (numerator zero, denominator nonzero).

• Vertical Tangent: Occurs where $\frac{dy}{dx}$ is undefined (denominator zero, numerator nonzero).

• Procedure:

  1. Set numerator or denominator of $\frac{dy}{dx}$ to zero as appropriate.

  2. Solve for $x$ or $y$.

  3. Substitute back into the original equation to find corresponding points.

• Geometric Intuition: For an ellipse, expect two horizontal and two vertical tangents, symmetrically arranged.

Example: For $\frac{dy}{dx} = \frac{8x - 24}{18y + 36}$:

• Horizontal tangents: $8x - 24 = 0 \implies x = 3$; substitute into original equation to find $y$.

• Vertical tangents: $18y + 36 = 0 \implies y = -2$; substitute into original equation to find $x$.

Second Derivative via Implicit Differentiation

To find the second derivative $\frac{d^2y}{dx^2}$ when $y$ is defined implicitly, differentiate $\frac{dy}{dx}$ with respect to $x$.

• Let $\frac{dy}{dx} = \frac{N(x, y)}{D(x, y)}$.

• Apply the quotient rule:

  • $\frac{d}{dx}\left(\frac{N}{D}\right) = \frac{N' \cdot D - N \cdot D'}{D^2}$

• When differentiating $N$ and $D$, apply the chain rule to $y$ terms (multiply by $\frac{dy}{dx}$).

• Substitute the expression for $\frac{dy}{dx}$ back into the result to eliminate $\frac{dy}{dx}$.

Example: If $\frac{dy}{dx} = \frac{8x - 24}{18y + 36}$, then

• $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{8x - 24}{18y + 36}\right)$

• Apply quotient and chain rules as above.

Logarithmic Differentiation

Logarithmic differentiation is useful for differentiating complicated expressions involving products, quotients, or variable exponents.

• Procedure:

  1. Take the natural logarithm of both sides: $\ln y = \ln(f(x))$.

  2. Use log properties to simplify: $\ln(a^b) = b \ln a$, $\ln(AB) = \ln A + \ln B$, $\ln(A/B) = \ln A - \ln B$.

  3. Differentiate both sides implicitly: $\frac{1}{y} \frac{dy}{dx} =$ derivative of right side.

  4. Solve for $\frac{dy}{dx}$, and substitute the original $y$ if needed.

• Advantage: Simplifies differentiation of complex expressions.

Example: For $y = x^x$:

• Take $\ln$ of both sides: $\ln y = x \ln x$

• Differentiate: $\frac{1}{y} \frac{dy}{dx} = \ln x + 1$

• Solve: $\frac{dy}{dx} = y(\ln x + 1) = x^x (\ln x + 1)$

Related Rates

Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another, often with respect to time.

• Variables: Let $A(t)$ be area, $s(t)$ be side length, $t$ be time.

• Relationship: $A = s^2$

• Differentiation: $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$

• Given: $\frac{ds}{dt} = 2$ in/min

• At $s = 5$ in: $\frac{dA}{dt} = 2 \times 5 \times 2 = 20$ in$^2$/min

• At $s = 7$ in: $\frac{dA}{dt} = 2 \times 7 \times 2 = 28$ in$^2$/min

• Insight: Linear growth in side length leads to accelerating area growth.

Key Terms and Definitions

• Implicit Equation: An equation that defines $y$ implicitly in terms of $x$.

• Implicit Differentiation: Differentiating both sides of an implicit equation with respect to $x$, using the chain rule for $y$-terms.

• Chain Rule: $\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$ if $y = y(x)$.

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

• Logarithmic Differentiation: Taking logarithms to simplify differentiation of products, quotients, and powers.

• Related Rates: Problems relating rates of change of related quantities via differentiation with respect to time.

Worked-Example Summary Table

Problem/Task	Key Steps
First derivative of $4x^2+24x+9y^2+36y=36$	Differentiate w.r.t $x$, apply chain rule to $y$-terms, solve for $\frac{dy}{dx}$
Horizontal tangents (slope 0)	Set numerator of $\frac{dy}{dx} = 0$, substitute into original equation to find points
Vertical tangents (undefined slope)	Set denominator of $\frac{dy}{dx} = 0$, substitute into original equation to find points
Second derivative from implicit $\frac{dy}{dx}$	Differentiate $\frac{dy}{dx}$ using quotient rule, apply chain rule, substitute $\frac{dy}{dx}$
Logarithmic differentiation (complicated $y$)	Take $\ln$ both sides, use log properties, differentiate implicitly, solve for $\frac{dy}{dx}$
Related rates (square area)	$A = s^2$; $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$; plug $s$ and $\frac{ds}{dt}$ to get $\frac{dA}{dt}$

Common Mistakes and Tips

• Do not write $\frac{dy}{dx}$ on both sides when you mean to apply $\frac{d}{dx}$ (operation vs. result).

• Always apply the chain rule to every differentiated $y$-term and include $\frac{dy}{dx}$.

• Factor $\frac{dy}{dx}$ early when solving for the derivative to simplify algebra.

• Use logarithmic differentiation to reduce complications from nested chain, quotient, or product rules.

• For related rates, clearly label dependent and independent variables and differentiate with respect to time.

Practice Problem

Differentiate implicitly: $\ln x \cdot \tan y = y^2$

• Use product rule on left: $\frac{d}{dx}(\ln x) \cdot \tan y + \ln x \cdot \frac{d}{dx}(\tan y)$

• Recall: $\frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx}$ and $\frac{d}{dx}(\ln x) = \frac{1}{x}$

• Move non-$\frac{dy}{dx}$ terms, factor $\frac{dy}{dx}$, and solve.

Action Items / Next Steps

• Practice implicit differentiation on various curves (circles, ellipses, higher-degree).

• Compute second derivatives for implicit curves and simplify by substituting $\frac{dy}{dx}$.

• Solve related rates problems, labeling all variables and units.

• Use logarithmic differentiation on functions with variable exponents or nested quotients.