Calc 1 Section 3.5: Implicit Differentiation

Welcome back to Professor Baker's Math Class! In this section, we're diving into the fascinating world of implicit differentiation. Unlike explicit functions where $y$ is directly defined in terms of $x$ (e.g., $y = x^2 + 1$), implicit functions define a relationship between $x$ and $y$ (e.g., $x^2 + y^2 = 25$). Our goal is to find $\frac{dy}{dx}$ when $y$ is not explicitly isolated.

Key Concepts and Techniques

Here's a breakdown of the implicit differentiation process:

  1. Differentiate both sides of the equation with respect to $x$. Remember that $y$ is a function of $x$, so you'll need to apply the chain rule when differentiating terms involving $y$.
  2. Apply the Chain Rule: Whenever you differentiate a term containing $y$ with respect to $x$, you must multiply by $\frac{dy}{dx}$. For example, $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.
  3. Solve for $\frac{dy}{dx}$: After differentiating, rearrange the equation to isolate $\frac{dy}{dx}$.

Example 1: Circle Equation

Let's consider the equation of a circle: $x^2 + y^2 = 25$. We want to find $\frac{dy}{dx}$.

Differentiating both sides with respect to $x$, we get:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$ $$2x + 2y \frac{dy}{dx} = 0$$

Now, solve for $\frac{dy}{dx}$:

$$2y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$$

So, $\frac{dy}{dx} = -\frac{x}{y}$. Notice that the derivative is expressed in terms of both $x$ and $y$.

Finding the Tangent Line

We can use implicit differentiation to find the equation of the tangent line to the circle at a specific point. For instance, at the point (3, 4):

$$\frac{dy}{dx} \Big|_{(3,4)} = -\frac{3}{4}$$

The slope of the tangent line at (3, 4) is $-\frac{3}{4}$. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, we get:

$$y - 4 = -\frac{3}{4}(x - 3)$$ $$y = -\frac{3}{4}x + \frac{9}{4} + 4$$ $$y = -\frac{3}{4}x + \frac{25}{4}$$

Example 2: A More Complex Curve

Let's tackle a slightly more challenging example: $x^3 + y^3 = 6xy$. This is known as the Folium of Descartes.

Differentiating both sides with respect to $x$:

$$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$$

Now, isolate $\frac{dy}{dx}$:

$$3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$$ $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$ $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$

Thus, $\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$.

Keep practicing, and you'll master implicit differentiation in no time! Remember to carefully apply the chain rule and solve for $\frac{dy}{dx}$. Good luck, and see you in the next section!