Implicit Differentiation

In calculus, we usually deal with explicit functions, where $y$ is isolated on one side of the equation (e.g., $y = 3x^2 + 5$). However, some equations define $y$ implicitly as a function of $x$, such as $x^2 + y^2 = 25$. When it is difficult or impossible to solve for $y$, we use implicit differentiation to find the derivative $\frac{dy}{dx}$.

The Core Concept

To find $\frac{dy}{dx}$ implicitly, we differentiate both sides of the equation with respect to $x$. Because $y$ is a function of $x$, we must apply the Chain Rule whenever we differentiate a term containing $y$. This means multiplying the derivative of the $y$ term by $\frac{dy}{dx}$.

Steps for Implicit Differentiation

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the standard derivative rules (Power, Product, Quotient). Whenever you differentiate a $y$ term, multiply it by $\frac{dy}{dx}$.
3. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation.
4. Factor out $\frac{dy}{dx}$.
5. Solve for $\frac{dy}{dx}$ by dividing both sides.

Example 1: Finding the Derivative

Problem: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$.

Solution:

1. Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$
2. Apply the power rule to $x^2$, and the chain rule to $y^2$: $2x + 2y \frac{dy}{dx} = 0$
3. Isolate the term with $\frac{dy}{dx}$: $2y \frac{dy}{dx} = -2x$
4. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$

Example 2: Finding a Tangent Line

Problem: Find the slope of the tangent line to $x^2 y + y^3 = 10$ at the point $(1, 2)$.

Solution:

1. Differentiate both sides. Use the Product Rule for $x^2 y$: $\frac{d}{dx}(x^2 y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(10)$ $\left(x^2 \frac{dy}{dx} + 2x \cdot y\right) + 3y^2 \frac{dy}{dx} = 0$
2. Move terms without $\frac{dy}{dx}$ to the right side: $x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -2xy$
3. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x^2 + 3y^2) = -2xy$
4. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}$
5. Plug in the point $(1, 2)$ to find the slope: $\frac{dy}{dx} = \frac{-2(1)(2)}{(1)^2 + 3(2)^2} = \frac{-4}{1 + 12} = -\frac{4}{13}$

The slope of the tangent line at $(1, 2)$ is $-\frac{4}{13}$.