Lecture 23: Implicit Differentiation

Motivation: Why Implicit Differentiation?

So far, you've learned to differentiate functions written explicitly — where one variable is expressed directly in terms of the other (e.g. $y = f(x)$).

But what if the equation doesn't solve nicely for $y$?
What if you're faced with a relation like:

$$x^2 + y^2 = 25$$

This is the equation of a circle — and notice, $y$ is not isolated!
In many real-world and geometric problems (like circles, ellipses, or strange curves), variables are intertwined in equations.

Implicit differentiation is the tool that lets us find slopes, tangents, and derivatives even when $y$ is not solved in terms of $x$.

The Idea Behind Implicit Differentiation

Suppose you have an equation involving both $x$ and $y$, and $y$ is a function of $x$ (even if not written explicitly).

To differentiate implicitly:

1. Differentiate both sides of the equation with respect to $x$.
2. When you differentiate a term involving $y$, use the chain rule:

   $$\frac{d}{dx}[y] = \frac{dy}{dx} \quad\text{and}\quad \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}$$

3. Solve for $\frac{dy}{dx}$ after taking derivatives.

Examples

Example 1: A Circle

$$x^2 + y^2 = 25$$

Differentiate both sides:

$$\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25] \Rightarrow 2x + 2y \cdot \frac{dy}{dx} = 0$$

Solve for $\frac{dy}{dx}$:

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

This gives the slope of the tangent to the circle at any point $(x, y)$.

Example 2: Exponential Function

Find $\frac{dy}{dx}$ if $x = e^y + y$

Solution:

Differentiate both sides:

$$\frac{d}{dx}[x] = \frac{d}{dx}[e^y + y]$$

$$1 = e^y \cdot \frac{dy}{dx} + \frac{dy}{dx}$$

Factor:

$$1 = \left(e^y + 1\right) \cdot \frac{dy}{dx} \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{e^y + 1}$$

Example 3: Logarithmic Function

Find $\frac{dy}{dx}$ if $\ln(xy) = y + x$

Solution:

Use the product rule inside the log:

$$\ln(xy) = \ln(x) + \ln(y)$$

Differentiate both sides:

$$\frac{d}{dx}[\ln(x) + \ln(y)] = \frac{d}{dx}[y + x]$$

$$\frac{1}{x} + \frac{1}{y} \cdot \frac{dy}{dx} = \frac{dy}{dx} + 1$$

Move all terms to one side:

$$\frac{1}{x} - 1 = \frac{dy}{dx} - \frac{1}{y} \cdot \frac{dy}{dx}$$

Factor $\frac{dy}{dx}$:

$$\frac{1}{x} - 1 = \frac{dy}{dx} \left(1 - \frac{1}{y}\right) \quad \Rightarrow \quad \frac{dy}{dx} = \frac{\frac{1}{x} - 1}{1 - \frac{1}{y}}$$

Summary of the Rules

Implicit Differentiation Procedure:

1. Differentiate both sides with respect to $x$.
2. Apply chain rule: whenever you differentiate $y$, multiply by $\frac{dy}{dx}$.
3. Rearrange and solve for $\frac{dy}{dx}$.

Exercises: Implicit Differentiation

Section 1: Problems without Natural Log or Exponential

Find $\frac{dy}{dx}$ for the following equations:

1. $x^2 + y^2 = 16$
2. $x^2 - y^2 = 9$
3. $xy = 1$
4. $x^3 + y^3 = 6xy$
5. $\sqrt{x} + \sqrt{y} = 1$
6. $\ln(x^2 + y^2) = 0$

Section 2: Problems Involving Exponentials

Find $\frac{dy}{dx}$ for the following equations:

1. $e^y + x^2 = y$
2. $x^x = y$
3. $x = y^y$
4. $e^{x^2y} = x^2 y$
5. $y = x^y$

Section 3: Problems Involving Natural Logarithms

Find $\frac{dy}{dx}$ for the following equations:

1. $\ln(xy) = x + y$
2. $x^2 + y^2 + \ln(xy) = 0$
3. $\ln(x) + \ln(y) = 1$
4. $\ln(x^2 y) = y$
5. $\ln(x + y) = xy$
6. $\ln(xy) + x = y$

Extra Advanced Problems

Find $\frac{dy}{dx}$ for the following equations:

1. $x^{y} + y^{x} = xy$

2. $\ln\big(x^2 + y^2\big) = xy + e^{xy}$

3. $e^{xy} + \ln(y) = x^2 y^3$

4. $x^{x^y} = y^{y^x}$

5. $\ln(y) = \frac{x + y}{x - y}$

6. $e^{x + y} = x^2 - y^2 + \ln(xy)$