Implicit Differentiation

Explicit vs. Implicit Functions

So far, most functions you've differentiated have been in an explicit form, where $y$ is isolated on one side of the equation, such as $y=3x^2+\sin (x)$. You are given an explicit recipe for finding $y$ from $x$.

However, many important curves are defined by an implicit relation, where $x$ and $y$ are mixed together. For example:

• $x^2+y^2=25$ (a circle)
• $x^3+y^3=6xy$ (a folium of Descartes)

For the circle, you could solve for $y$ to get two separate explicit functions, $y=\sqrt{25-x^2}$ (the top semicircle) and $y=-\sqrt{25-x^2}$ (the bottom semicircle). But for more complex relations, like the folium of Descartes, solving for $y$ can be extremely difficult or even impossible using standard algebraic methods.

Professor Leonard provides an excellent introduction distinguishing between explicit and implicit forms and explains why we need a new method.

The Core Technique of Implicit Differentiation

Fortunately, we don't need to solve for $y$ to find its derivative. The method of implicit differentiation consists of three main steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Crucially, whenever you differentiate a term involving $y$, you must treat $y$ as a function of $x$ (i.e., $y=y(x)$) and apply the Chain Rule. This means you will multiply by $\frac{dy}{dx}$ (or $y^{\prime }$).
3. Solve the resulting equation algebraically for $\frac{dy}{dx}$.

Let's break down Step 2, as it's the heart of the process. If you have a term like $y^3$, you are differentiating it with respect to $x$. From our last lesson, you know this requires the Chain Rule:

Notice the $\frac{dy}{dx}$ term that appears. This happens every time you differentiate an expression involving $y$. For differentiating an x term like x^3 with respect to x, the chain rule would technically give $3x^2\cdot \frac{dx}{dx}$, but since $\frac{dx}{dx}=1$, we usually omit it.

The following video segment explains this core principle with exceptional clarity, directly linking it to the Chain Rule.

A Step-by-Step Example: The Circle

Let's apply this process to our circle equation, $x^2+y^2=25$, to find $\frac{dy}{dx}$.

Let's review the steps from the reading:

1. Differentiate both sides with respect to x:

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

2. Apply the Sum Rule and Chain Rule:
   The derivative of $x^2$ with respect to $x$ is $2x$.
   The derivative of $y^2$ with respect to $x$ requires the Chain Rule: $2y\cdot \frac{dy}{dx}$.
   The derivative of the constant 25 is 0.
   This gives us:

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

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

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

Notice that the derivative is expressed in terms of both $x$ and $y$. This is typical for implicit differentiation. To find the slope at a specific point, you need both coordinates. For example, at the point (3, -4) on the circle (from the image earlier), the slope is:

Handling More Complex Relations

The true power of this technique is for relations where other rules, like the Product Rule, are also needed. Let's consider the folium of Descartes.

Let's find the derivative for $x^3+y^3=6xy$.

Let's recap the key step from that example. When differentiating the right side, $6xy$, you must use the Product Rule, treating 6x as the first function and y as the second:

This is a common place for errors, so it's vital to correctly identify products of functions of $x$ and $y$.

For a visual walkthrough of more examples involving the product rule and trigonometric functions, this video is very helpful.

Finding the Second Derivative

You can also find higher-order derivatives implicitly. To find $y^{\prime \prime }$ (or $\frac{d^{2}y}{d{x}^2}$), you simply differentiate your expression for $y^{\prime }$ with respect to $x$. This process usually involves the Quotient or Product Rule and one extra substitution step.

Let's find $y^{\prime \prime }$ for $x^4+y^4=16$.

The key steps demonstrated in the text are:

1. First, find $y^{\prime }$ using implicit differentiation. For $x^4+y^4=16$, this gives $y^{\prime }=-\frac{x^3}{y^3}$.
2. Next, differentiate $y^{\prime }$ with respect to $x$ using the Quotient Rule. Remember that $y$ is a function of $x$, so when you differentiate a $y$ term, a $y^{\prime }$ will appear. $$y^{\prime \prime }=\frac{d}{dx}\left(-\frac{x^3}{y^3}\right)=-\frac{(3x^2)(y^3)-(x^3)(3y^2{y}^{\prime })}{(y^3{)}^2}$$
3. The expression for $y^{\prime \prime }$ now contains a $y^{\prime }$. Substitute your original expression for $y^{\prime }$ back into this equation. $$y^{\prime \prime }=-\frac{3x^2{y}^3-3x^3{y}^2\left(-\frac{x^3}{y^3}\right)}{y^6}$$
4. Simplify the expression algebraically. As shown in the text, you can often use the original equation ($x^4+y^4=16$) to simplify the final result further.

Conclusion

Today we've added a crucial technique to our calculus toolkit. Implicit differentiation frees us from the constraint of needing functions in explicit form. By systematically applying the Chain Rule, we can find the rate of change for a vast array of implicitly defined curves.

Key Takeaways:

• Use implicit differentiation when a relation is given where $y$ is not explicitly solved as a function of $x$.
• The core of the method is to differentiate both sides of the equation with respect to $x$.
• Whenever you differentiate a term involving $y$, you are using the Chain Rule, which means you must multiply that term's derivative by $\frac{dy}{dx}$.
• After differentiating all terms, use algebra to isolate $\frac{dy}{dx}$.
• The resulting derivative, which may contain both $x$ and $y$, gives the slope of the tangent line at any point $(x,y)$ on the curve.

In our next lesson, we will begin using our powerful new differentiation tools to analyze the behavior of functions. We will start with Rolle's Theorem and the Mean Value Theorem, which provide fundamental insights into how a function's average rate of change relates to its instantaneous rate of change.