2.7 Implicit Differentiation

Explicit and Implicit Functions

• Explicit Form: One variable is explicitly solved in terms of the other, such as $y = f(x)$. Example: $y = x^2 + 3$.

• Implicit Form: The relationship between variables is implied by an equation where $y$ is not isolated. Example: $x^2 + y^2 = 25$ or $x^3 + y^3 = 6xy$.

• Implicit Differentiation: A procedure used to find derivatives when it is difficult or impossible to solve for $y$ as a function of $x$ explicitly.

Applying the Chain Rule

• Implicit differentiation requires differentiating with respect to ($wrt$) $x$.

• When differentiating terms involving $y$, you must assume $y$ is a differentiable function of $x$ and apply the Chain Rule, which introduces a $\frac{dy}{dx}$ factor.

• Examples of Differentiation $wrt \text{ } x$:

  • Terms with only $x$: $\frac{d}{dx}[x^3] = 3x^2$.

  • Terms with only $y$: $\frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}$.

  • Combined terms (Sum Rule): $\frac{d}{dx}[x + 3y] = 1 + 3\frac{dy}{dx}$.

  • Combined terms (Product Rule): $\frac{d}{dx}[xy^2] = x(2y\frac{dy}{dx}) + y^2(1) = 2xy\frac{dy}{dx} + y^2$.

Procedural Steps for Implicit Differentiation

1. Differentiate both sides of the equation with respect to $x$.

2. Collect all terms involving $\frac{dy}{dx}$ on the left side and move all other terms to the right side.

3. Factor $\frac{dy}{dx}$ out of the left side.

4. Solve for $\frac{dy}{dx}$ by dividing.

• Example 3 Calculation: For the equation $y^3 + y^2 - 5y - x^2 = -4$, differentiating yields $3y^2 \frac{dy}{dx} + 2y \frac{dy}{dx} - 5 \frac{dy}{dx} - 2x = 0$. Solving for the derivative results in $\frac{dy}{dx} = \frac{2x}{3y^2 + 2y - 5}$.

Finding Slopes and Tangent Lines

• The derivative $\frac{dy}{dx}$ provides the slope of the tangent line at any point $(x, y)$ on the graph.

• Slopes at Specific Points:

  • In the graph of $y^3 + y^2 - 5y - x^2 = -4$ (Figure 2.31), the slope at $(1, -3)$ is $\frac{2(1)}{3(-3)^2 + 2(-3) - 5} = \frac{2}{16} = \frac{1}{8}$.

  • Example 5: For the hyperbola $x^2 - y^2 = 16$, the slope at point $(1, 1)$ found implicitly is $\frac{dy}{dx} = 2$.

Application: Demand Functions

• Implicit differentiation can be used to determine the rate of change in economic models.

• Example 6: Given the demand function $p = \frac{100}{0.0001x^2 + 0.01x + 1}$, where $p$ is price and $x$ is demand (in thousands of units).

• To find the rate of change of demand with respect to price ($\frac{dx}{dp}$) at $x = 100$:

  • Differentiate implicitly to find $\frac{dx}{dp} = -75$.

  • Interpretation: When $x = 100$, the demand drops at a rate of $75$ thousand units for every $\$1$ increase in price.