Welcome back to Professor Baker's Math Class! Today's focus is on implicit differentiation, a powerful tool for finding derivatives when $y$ is not explicitly defined as a function of $x$. Get ready to dive into some fascinating examples and sharpen your calculus skills!

What is Implicit Differentiation?

Sometimes, we encounter equations where $y$ is intertwined with $x$, making it difficult or impossible to isolate $y$ and express it in the form $y = f(x)$. In these cases, we use implicit differentiation. The key idea is to differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule whenever we differentiate a term involving $y$.

Examples and Applications

Example 1: Finding $\frac{dy}{dx}$ for $x^2 + y^2 = 25$

Let's consider the equation of a circle: $x^2 + y^2 = 25$. To find $\frac{dy}{dx}$, we differentiate both sides with respect to $x$:

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

$$2x + 2y \frac{dy}{dx} = 0$$\

Solving for $\frac{dy}{dx}$, we get:

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

Example 2: Tangent to the Folium of Descartes

The Folium of Descartes is a classic example where implicit differentiation shines. Its equation is $x^3 + y^3 = 6xy$. Let's find $\frac{dy}{dx}$:

$$\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)$$\

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

Rearranging and solving for $\frac{dy}{dx}$:

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

Now, let's find the tangent line at the point (3, 3). Plugging in $x = 3$ and $y = 3$ into our expression for $\frac{dy}{dx}$:

$$\frac{dy}{dx}|_{(3,3)} = \frac{2(3) - (3)^2}{(3)^2 - 2(3)} = \frac{6 - 9}{9 - 6} = -1$$\

So, the slope of the tangent line at (3, 3) is -1. The equation of the tangent line is:

$y - 3 = -1(x - 3)$, which simplifies to $y = -x + 6$.

Example 3: Find $y'$ if $\sin(x + y) = y^2 \cos(x)$

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

$$\cos(x+y)(1+y') = 2yy'\cos(x) - y^2\sin(x)$$\

Rearranging and solving for $y'$:

$$y' = \frac{-\cos(x+y) - y^2 \sin(x)}{\cos(x+y)-2y \cos(x)}$$\

Key Takeaways

  • Implicit differentiation allows us to find derivatives of implicitly defined functions.
  • Remember to apply the chain rule when differentiating terms involving $y$.
  • Implicit differentiation is useful for finding tangent lines to curves defined by implicit equations.

Keep practicing, and you'll become a master of implicit differentiation! Don't hesitate to ask questions in the comments below. Good luck with your studies!