Grade 12

Advanced

9 min

Lesson 3 of 12

The Chain Rule and Implicit Differentiation

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Learn to differentiate composite functions and handle equations where the dependent variable is not isolated.

Imagine you are tracking a viral outbreak: the number of infections depends on social interactions, which in turn depends on the time of day. How do you calculate the rate of change of infections relative to time when the variables are nested like Russian dolls?

Learning Objectives

1.Apply the Chain Rule to differentiate composite functions by identifying 'inner' and 'outer' layers
2.Perform implicit differentiation on equations where the dependent variable cannot be easily isolated
3.Calculate the slope of a tangent line for non-function curves like circles and ellipses

The Chain Rule: Peeling the Onion

In calculus, we often encounter composite functions, written as

f(g(x))

. To differentiate these, we use the Chain Rule. Think of a composite function like an onion: to get to the center, you must peel the outer layer first. The rule states that the derivative of the whole is the derivative of the outside function (evaluated at the inside) multiplied by the derivative of the inside function. Formally, if

y = f(u)

and

u = g(x)

, then:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

This allows us to break complex expressions into manageable 'chunks.' We prioritize identifying the inner function

g(x)

first, then apply the power rule or trigonometric rules to the outer shell

f(u)

Find the derivative of $y = (4x^3 + 2)^5$ .

1. Identify the inner function: $u = 4x^3 + 2$ . Its derivative is $\frac{du}{dx} = 12x^2$ . 2. Identify the outer function: $y = u^5$ . Its derivative is $\frac{dy}{du} = 5u^4$ . 3. Multiply them: $\frac{dy}{dx} = 5(u)^4 \cdot 12x^2$ . 4. Substitute $u$ back: $\frac{dy}{dx} = 5(4x^3 + 2)^4 \cdot 12x^2 = 60x^2(4x^3 + 2)^4$ .

Quick Check

If $y = \sin(x^2)$ , what is the 'inner' function $g(x)$ and its derivative?

Look at what is inside the parentheses of the sine function.

Answer

The inner function is $x^2$ and its derivative is $2x$ .

Implicit Differentiation: When y is Hidden

Most functions we've seen are explicit, like $y = x^2$ . But what about $x^2 + y^2 = 25$ ? This is a circle, and $y$ isn't isolated. Implicit differentiation allows us to find the derivative $\frac{dy}{dx}$ without solving for $y$ first. The secret is treating $y$ as a function of $x$ , $y(x)$ . Whenever you differentiate a term containing $y$ , you must apply the Chain Rule, which results in a $\frac{dy}{dx}$ term appearing. For example, the derivative of $y^2$ with respect to $x$ is $2y \cdot \frac{dy}{dx}$ . Once you differentiate every term on both sides of the equation, you simply use algebra to isolate $\frac{dy}{dx}$ .

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

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

Quick Check

When differentiating $y^3$ implicitly with respect to $x$ , what is the result?

Use the power rule on

y

, then multiply by the derivative of the 'inside' (which is

y

Answer

$3y^2 \frac{dy}{dx}$

Advanced Curves and Tangents

Implicit differentiation is essential for analyzing complex geometric shapes like ellipses or hyperbolas. In these cases, a single $x$ -value might correspond to multiple $y$ -values. To find the slope of a tangent line at a specific point $(x, y)$ , we differentiate implicitly and then plug in both coordinates. This is a powerful tool in physics and engineering, where relationships between variables are often defined by constraints (like the path of a planet) rather than direct assignments. Remember: if you see a product like $xy$ , you must use the Product Rule combined with implicit differentiation: $\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx}$ .

Find the slope of the tangent line to $x^3 + y^3 = 6xy$ at the point $(3, 3)$ .

1. Differentiate implicitly: $3x^2 + 3y^2\frac{dy}{dx} = 6(1 \cdot y + x \cdot \frac{dy}{dx})$ . 2. Distribute the 6: $3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$ . 3. Plug in $x=3, y=3$ : $3(3)^2 + 3(3)^2\frac{dy}{dx} = 6(3) + 6(3)\frac{dy}{dx}$ . 4. Simplify: $27 + 27\frac{dy}{dx} = 18 + 18\frac{dy}{dx}$ . 5. Solve for $\frac{dy}{dx}$ : $9\frac{dy}{dx} = -9$ , so $\frac{dy}{dx} = -1$ .

Key Takeaways

1The Chain Rule is used for nested functions: differentiate the outside, keep the inside the same, then multiply by the derivative of the inside.
2Implicit differentiation is required when $y$ cannot be easily isolated; always multiply by $\frac{dy}{dx}$ when differentiating $y$ terms.
3The Product Rule ( $uv' + vu'$ ) must be applied to terms like $xy$ during implicit differentiation.
4The derivative $\frac{dy}{dx}$ represents the slope of the tangent line, even for non-function curves.

Quick Check

Multiple Choice

What is the derivative of $y = \cos(5x)$ ?

Multiple Choice

Differentiate $y^2 = x$ implicitly with respect to $x$ .

True/False

When using implicit differentiation on the term $5y$ , the derivative with respect to $x$ is simply $5$ .

What's Next?

Review Tomorrow

In 24 hours, try to explain the 'Onion Analogy' for the Chain Rule to someone else and write down the implicit derivative of $x^2 + y^2 = r^2$ from memory.

Practice Activity

Find the equation of the tangent line to the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ at the point $(\sqrt{2}, \frac{3}{\sqrt{2}})$ .

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The Chain Rule and Implicit Differentiation

Learning Objectives

The Chain Rule: Peeling the Onion

Implicit Differentiation: When y is Hidden

Advanced Curves and Tangents

Key Takeaways

Quick Check

What's Next?

The Chain Rule and Implicit Differentiation

Learning Objectives

The Chain Rule: Peeling the Onion

Implicit Differentiation: When y is Hidden

Advanced Curves and Tangents

Key Takeaways

Quick Check

What's Next?