1) Introduction: The Power of Composition - Need for the Chain Rule
So far, we've learned rules to differentiate sums, differences, products, and quotients of functions, as well as powers of \(x\). However, we often encounter functions that are compositions of other functions. A composite function is essentially a function "inside" another function, like \( \sin(x^2) \) or \( \sqrt{3x + 1} \).
For composite functions, our previous rules fail directly. We need a new, powerful rule called the Chain Rule. The Chain Rule is arguably one of the most important differentiation rules in calculus. It allows us to differentiate composite functions systematically. We'll also explore how the Chain Rule is crucial for implicit differentiation, which deals with relationships between \(x\) and \(y\) that are not explicitly solved for \(y\) as a function of \(x\).
What is a Composite Function? (Review)
A composite function is formed when one function is plugged into another. If we have functions \( f \) and \( g \), the composition \( f(g(x)) \) (read as "f of g of x") means we first apply \( g \) to \( x \), and then apply \( f \) to the result \( g(x) \). For example, if \( f(u) = \sin(u) \) and \( g(x) = x^2 \), then \( f(g(x)) = \sin(x^2) \). Here, \( g(x) = x^2 \) is the "inner function" and \( f(u) = \sin(u) \) is the "outer function."
2) The Chain Rule: Differentiating Composite Functions
The Chain Rule provides a formula for finding the derivative of a composite function \( f(g(x)) \).
Rule 4.1: Chain Rule
If \( g(x) \) is differentiable at \( x \) and \( f(u) \) is differentiable at \( u = g(x) \), then the composite function \( F(x) = f(g(x)) \) is differentiable at \( x \), and its derivative is given by:
\( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
Alternatively, if we let \( u = g(x) \) and \( y = f(u) = f(g(x)) \), then we can write the Chain Rule in Leibniz notation as:
\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)
In words: To differentiate a composite function \( f(g(x)) \), you take the derivative of the "outer function" \( f \) evaluated at the "inner function" \( g(x) \), and then multiply by the derivative of the "inner function" \( g'(x) \). Think of it as differentiating "outside" then multiplying by the derivative of the "inside." In Leibniz notation, the Chain Rule expresses the rate of change of \( y \) with respect to \( x \) as the product of the rate of change of \( y \) with respect to the intermediate variable \( u \), and the rate of change of \( u \) with respect to \( x \).
Examples of the Chain Rule
- Example 1: \( y = \sin(x^2) \)
Identify the outer and inner functions:
- Outer function: \( f(u) = \sin(u) \), so \( f'(u) = \cos(u) \)
- Inner function: \( g(x) = x^2 \), so \( g'(x) = 2x \)
- Example 2: \( f(x) = (3x + 1)^4 \)
Identify the outer and inner functions:
- Outer function: \( f(u) = u^4 \), so \( f'(u) = 4u^3 \)
- Inner function: \( g(x) = 3x + 1 \), so \( g'(x) = 3 \)
- Example 3: \( y = \sqrt{x^2 + 5} \)
Rewrite the square root as a power: \( y = (x^2 + 5)^{1/2} \).
Identify the outer and inner functions:
- Outer function: \( f(u) = u^{1/2} \), so \( f'(u) = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \)
- Inner function: \( g(x) = x^2 + 5 \), so \( g'(x) = 2x \)
- Example 4: \( h(x) = \frac{1}{(2x - 3)^2} \)
Rewrite using a negative exponent: \( h(x) = (2x - 3)^{-2} \).
Identify the outer and inner functions:
- Outer function: \( f(u) = u^{-2} \), so \( f'(u) = -2u^{-3} \)
- Inner function: \( g(x) = 2x - 3 \), so \( g'(x) = 2 \)
3) Implicit Differentiation: Derivatives of Implicitly Defined Functions
Sometimes, relationships between \( x \) and \( y \) are given by equations where \( y \) is not explicitly isolated as a function of \( x \). For example, \( x^2 + y^2 = 25 \) or \( x^3 + y^3 = 6xy \). These are called implicit functions (or more accurately, implicitly defined relations). Implicit differentiation is a technique to find \( \frac{dy}{dx} \) directly from such implicit equations.
The key idea in implicit differentiation is to treat \( y \) as a function of \( x \) and use the Chain Rule whenever we differentiate a term involving \( y \) with respect to \( x \). Because \( y \) is a function of \( x \), differentiating \( y \) with respect to \( x \) is just \( \frac{dy}{dx} \) (or \( y' \)). But differentiating something like \( y^2 \) or \( \sin(y) \) with respect to \( x \) requires the Chain Rule because we have a function of \( y \) and \( y \) itself is a function of \( x \).
Implicit vs. Explicit Functions
- Explicit Function: \( y \) is given directly in terms of \( x \), like \( y = x^2 + 3x \) or \( f(x) = \sqrt{x} \). We can directly substitute a value of \( x \) to find \( y \).
- Implicit Function (Relation): Relationship between \( x \) and \( y \) is given by an equation, but \( y \) is not isolated, like \( x^2 + y^2 = 25 \) or \( xy + \sin(y) = x \). We can't easily (or sometimes at all) solve for \( y \) explicitly as \( y = f(x) \).
4) Steps for Implicit Differentiation
Here are the steps to perform implicit differentiation:
- Differentiate both sides of the equation with respect to \( x \). Remember to treat \( y \) as a function of \( x \).
- Apply differentiation rules as needed. This will often involve the Chain Rule when differentiating terms involving \( y \). Specifically, if you have a term involving \( y^n \), its derivative with respect to \( x \) is \( \frac{d}{dx}[y^n] = n y^{n-1} \cdot \frac{dy}{dx} \) (using Chain Rule with outer function \( u^n \) and inner function \( y = y(x) \)). Similarly, for \( \sin(y) \), the derivative is \( \cos(y) \cdot \frac{dy}{dx} \), and so on.
- Collect all terms containing \( \frac{dy}{dx} \) (or \( y' \)) on one side of the equation and all other terms on the other side.
- Factor out \( \frac{dy}{dx} \) from the terms you collected.
- Solve for \( \frac{dy}{dx} \) by dividing both sides by the factor you factored out. This will give you \( \frac{dy}{dx} \) in terms of \( x \) and \( y \).
5) Examples of Implicit Differentiation
Example 1: Find \( \frac{dy}{dx} \) for \( x^2 + y^2 = 25 \) (Equation of a circle)
- Differentiate both sides with respect to \( x \): \( \frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25] \)
- Apply differentiation rules: \( \frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25] \) \( 2x + 2y \cdot \frac{dy}{dx} = 0 \) (Using Power Rule for \( x^2 \), Chain Rule for \( y^2 \) treating \( y \) as \( y(x) \), and Constant Rule for 25).
- Collect \( \frac{dy}{dx} \) terms on one side: \( 2y \frac{dy}{dx} = -2x \)
- Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y} \)
So, for \( x^2 + y^2 = 25 \), we have \( \frac{dy}{dx} = -\frac{x}{y} \). Notice that the derivative \( \frac{dy}{dx} \) is expressed in terms of both \( x \) and \( y \), which is typical in implicit differentiation.
Example 2: Find \( y' \) for \( x^3 + y^3 = 6xy \) (Folium of Descartes curve)
- Differentiate both sides with respect to \( x \): \( \frac{d}{dx}[x^3 + y^3] = \frac{d}{dx}[6xy] \)
- Apply differentiation rules: \( \frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = 6 \frac{d}{dx}[xy] \) (Constant Multiple Rule on right side) \( 3x^2 + 3y^2 \cdot \frac{dy}{dx} = 6 \left( \frac{d}{dx}[x] \cdot y + x \cdot \frac{d}{dx}[y] \right) \) (Power and Chain Rules on left, Product Rule on right) \( 3x^2 + 3y^2 \frac{dy}{dx} = 6 \left( 1 \cdot y + x \cdot \frac{dy}{dx} \right) \) (Derivative of \( x \) is 1, derivative of \( y \) is \( \frac{dy}{dx} \)) \( 3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx} \) (Distribute the 6)
- Collect \( \frac{dy}{dx} \) terms on one side: \( 3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2 \)
- Factor out \( \frac{dy}{dx} \): \( \frac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2 \)
- Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} \)
We can simplify this by dividing numerator and denominator by 3: \( \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} \). So, for \( x^3 + y^3 = 6xy \), we have \( \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} \).
Example 3: Implicit Differentiation with Trigonometric Function: \( \cos(y) = x^2y \)
- Differentiate both sides with respect to \( x \): \( \frac{d}{dx}[\cos(y)] = \frac{d}{dx}[x^2y] \)
- Apply differentiation rules: \( -\sin(y) \cdot \frac{dy}{dx} = \frac{d}{dx}[x^2] \cdot y + x^2 \cdot \frac{d}{dx}[y] \) (Chain Rule on left, Product Rule on right) \( -\sin(y) \frac{dy}{dx} = (2x) \cdot y + x^2 \cdot \frac{dy}{dx} \) \( -\sin(y) \frac{dy}{dx} = 2xy + x^2 \frac{dy}{dx} \)
- Collect \( \frac{dy}{dx} \) terms on one side: \( -\sin(y) \frac{dy}{dx} - x^2 \frac{dy}{dx} = 2xy \)
- Factor out \( \frac{dy}{dx} \): \( \frac{dy}{dx} (-\sin(y) - x^2) = 2xy \)
- Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{2xy}{-\sin(y) - x^2} = -\frac{2xy}{\sin(y) + x^2} \)
Thus, for \( \cos(y) = x^2y \), we get \( \frac{dy}{dx} = -\frac{2xy}{\sin(y) + x^2} \).
6) Note Box: Chain Rule and Implicit Differentiation - The Connection
Chain Rule is the Key to Implicit Differentiation
Implicit differentiation fundamentally relies on the Chain Rule. When we differentiate a term involving \( y \) with respect to \( x \), we are using the Chain Rule because \( y \) is understood to be a function of \( x \).
For example, when we differentiated \( y^2 \) with respect to \( x \) in \( x^2 + y^2 = 25 \), we used the Chain Rule:
\( \frac{d}{dx}[y^2] = \frac{d}{dy}[y^2] \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx} \)
Here, the "outer function" is \( f(u) = u^2 \) and the "inner function" is \( u = y = y(x) \). So, the Chain Rule is not a separate rule for implicit differentiation; it's the underlying principle that makes implicit differentiation work.
7) Practice Questions - Chain Rule and Implicit Differentiation
Practice applying the Chain Rule and Implicit Differentiation techniques.
Fundamental Practice Questions (Chain Rule)
Instructions: Find the derivative \( f'(x) \) or \( \frac{dy}{dx} \) for each function using the Chain Rule.
- Q1. \( y = (4x + 3)^5 \)
- Q2. \( f(x) = \sin(3x) \)
- Q3. \( g(x) = \cos(x^2 + 1) \)
- Q4. \( y = \sqrt{5x - 2} \) (Rewrite as \( (5x - 2)^{1/2} \))
- Q5. \( h(x) = (x^2 - 2x + 5)^{-3} \)
- Q6. \( y = (\frac{1}{x + 1})^2 \) (Rewrite as \( (x + 1)^{-2} \))
- Q7. \( f(x) = (x^3 - 2)^4 \)
- Q8. \( g(x) = \sin^2(x) \) (This means \( (\sin(x))^2 \). Outer function is \( u^2 \), inner is \( \sin(x) \))
- Q9. \( y = \cos^3(x) \) (This means \( (\cos(x))^3 \))
- Q10. \( h(x) = \sqrt{x^3 + 2x} \)
Fundamental Practice Questions (Implicit Differentiation)
Instructions: Find \( \frac{dy}{dx} \) using implicit differentiation for each equation.
- Q1. \( x^2 + y^2 = 36 \)
- Q2. \( x^3 - y^3 = 8 \)
- Q3. \( xy = 4 \)
- Q4. \( x^2 + xy + y^2 = 7 \)
- Q5. \( x^2 = y^3 \)
- Q6. \( y^2 - 2x = 1 \)
- Q7. \( x = y^2 + 1 \)
- Q8. \( \sin(y) = x \)
- Q9. \( \cos(y) = x^2 \)
- Q10. \( xy + y^2 = x \)
Challenging Practice Questions
Instructions: These problems may require combining Chain Rule and Implicit Differentiation, or more complex algebra and conceptual understanding.
- Q1. Find the equation of the tangent line to the curve \( x^2 + y^2 = 25 \) at the point \( (3, 4) \). (Use implicit differentiation to find the slope \( \frac{dy}{dx} \) at \( (3, 4) \)).
- Q2. Find \( \frac{dy}{dx} \) for \( \sin(xy) = x \). (Requires Chain Rule and Product Rule within implicit differentiation).
- Q3. Find \( \frac{dy}{dx} \) for \( \sqrt{x} + \sqrt{y} = 4 \).
- Q4. Find \( \frac{dy}{dx} \) for \( y = \sin((x^2 + 1)^3) \). (Nested Chain Rule - apply Chain Rule multiple times).
- Q5. For the curve \( x^2 + xy + y^2 = 7 \), find the points where the tangent line is horizontal. (Find \( \frac{dy}{dx} \), set it to 0, and solve for \( x \) and \( y \) that satisfy both \( \frac{dy}{dx} = 0 \) and the original equation).
8) Summary & Cheat Sheet - Chain Rule and Implicit Differentiation
Let's summarize the key concepts for Chain Rule and Implicit Differentiation.
8.1) Chain Rule
For composite function \( f(g(x)) \): \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \) or \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) (if \( y = f(u), u = g(x) \)).
"Derivative of outside (evaluated at inside) times derivative of inside."
8.2) Implicit Differentiation - Steps
- Differentiate both sides of implicit equation with respect to \( x \).
- Use Chain Rule for \( y \) terms: \( \frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx} \), etc.
- Collect terms with \( \frac{dy}{dx} \) on one side, others on the other.
- Factor out \( \frac{dy}{dx} \).
- Solve for \( \frac{dy}{dx} \). Answer may contain both \( x \) and \( y \).
8.3) Key Connections
- Chain Rule is essential for differentiating composite functions.
- Implicit differentiation uses Chain Rule to differentiate equations where \( y \) is not explicitly solved for in terms of \( x \).
- In implicit differentiation, treat \( y \) as a function of \( x \) and apply Chain Rule accordingly.
Fantastic! You've now unlocked the power of the Chain Rule and Implicit Differentiation. These are critical techniques for handling a vast array of functions and relations in calculus and beyond. Keep practicing to master them! Next, we'll move on to Higher-Order Derivatives.