3.1 The Chain Rule

When differentiating a composite function, we use the Chain Rule. The Chain Rule states that if you have a composite function \( f(g(x)) \), its derivative is:

In other words, to find the derivative of a composite function:

1. Differentiate the outer function \( f \) with respect to its input \( g(x) \), keeping \( g(x) \) as it is.
2. Multiply by the derivative of the inner function \( g(x) \) with respect to \( x \).

The Chain Rule is essential in calculus because it allows us to handle the derivatives of complex, nested functions by breaking them down step-by-step.

Chain Rule for Three Functions

When differentiating a composite function of three functions, we can extend the Chain Rule to handle the additional layer. Suppose we have three functions \( f(x) \), \( g(x) \), and \( h(x) \), and we're dealing with the composite function \( f(g(h(x))) \).

Explanation

To differentiate a composite function with three functions \( f(g(h(x))) \), we apply the following steps:

1. Differentiate the outermost function \( f \) with respect to its input \( g(h(x)) \), keeping \( g(h(x)) \) as it is.
2. Multiply by the derivative of the middle function \( g \) with respect to its input \( h(x) \), keeping \( h(x) \) as it is.
3. Multiply by the derivative of the innermost function \( h \) with respect to \( x \).

This rule allows us to tackle complex nested functions by working through each function layer by layer.

Example

\( f(g(h(x))) = \cos(\sin(5x)) \), where:

• \( f(x) = \cos(x) \)
• \( g(x) = \sin(x) \)
• \( h(x) = 5x \)

We find each derivative step-by-step:

1. Differentiate \( f(x) = \cos(x) \) with respect to \( x \):
   \[ f'(x) = -\sin(x) \]
   So, \( f'(g(h(x))) = -\sin(\sin(5x)) \).
2. Differentiate \( g(x) = \sin(x) \) with respect to \( x \):
   \[ g'(x) = \cos(x) \]
   So, \( g'(h(x)) = \cos(5x) \).
3. Differentiate \( h(x) = 5x \) with respect to \( x \):
   \[ h'(x) = 5 \]

Putting it all together, we get:

Or simplifying:

This is the derivative of the composite function \( \cos(\sin(5x)) \) using the Chain Rule for three functions.

3.2 Implicit Differentiation

Explicit equations

define one variable in terms of another directly. For example, in the explicit equation \[ y = x^2 + 4x + 7 \] , \( y \) is given directly in terms of \( x \).

Implicit equations

represent a relationship between variables that isn't explicitly solved for one variable. For example, \[ e^x + y \cdot \ln(y) = 10 \] does not isolate \( y \) in terms of \( x \). Implicit differentiation is often used to find the derivative in cases like this.

Chain Rule and Implicit Differentiation

To differentiate implicit equations, we often need the

chain rule

. The chain rule states that if we are differentiating a function that involves another function (like \( y \) with respect to \( x \)), we need to account for that by including \( \frac{dy}{dx} \).

• Differentiating in terms of \( x \):

  • For simple expressions like \( x \), \( \frac{d}{dx} x = 1 \).
  • For powers of \( x \), like \( x^2 \), we use the power rule: \( \frac{d}{dx} x^2 = 2x \).
  • For functions of \( x \) inside another function, like \( e^{5x} \), we apply the chain rule: \( \frac{d}{dx} e^{5x} = e^{5x} \cdot 5 \) (since the derivative of \( 5x \) is \( 5 \)).

• Differentiating in terms of \( y \) when \( y \) is implicitly a function of \( x \):

  • When differentiating \( y \), we treat it as a function of \( x \). So, \( \frac{d}{dx} y = \frac{dy}{dx} \).
  • For powers of \( y \), like \( y^2 \), we use the chain rule: \( \frac{d}{dx} y^2 = 2y \cdot \frac{dy}{dx} \).
  • For more complex expressions like \( e^{5y} \), we also use the chain rule: \( \frac{d}{dx} e^{5y} = e^{5y} \cdot 5 \cdot \frac{dy}{dx} \).

Steps for Implicit Differentiation

To solve for \( \frac{dy}{dx} \) in an implicit equation, we follow these steps:

1. Take the derivative

   of both sides of the equation with respect to \( x \), applying the chain rule whenever you differentiate a term involving \( y \).

2. Gather all terms involving \( \frac{dy}{dx} \)

   on one side of the equation. Move other terms to the opposite side.

3. Factor out \( \frac{dy}{dx} \)

   on the side where all \( \frac{dy}{dx} \) terms are collected.

4. Solve for \( \frac{dy}{dx} \)

   by isolating it on one side of the equation.

Example: Differentiating \( y^2 - 5x^3 = 3y \)

Let's use the steps above to differentiate the implicit equation \( y^2 - 5x^3 = 3y \).

1. Differentiate each term with respect to \( x \):

   • For \( y^2 \): Using the chain rule, \( \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} \).
   • For \( -5x^3 \): This is straightforward differentiation, so \( \frac{d}{dx}(-5x^3) = -15x^2 \).
   • For \( 3y \): Using the chain rule again, \( \frac{d}{dx}(3y) = 3 \cdot \frac{dy}{dx} \).

   Putting it all together:

   \[ 2y \cdot \frac{dy}{dx} - 15x^2 = 3 \cdot \frac{dy}{dx}. \]

2. Collect all \( \frac{dy}{dx} \) terms on the left side:

   \[ 2y \cdot \frac{dy}{dx} - 3 \cdot \frac{dy}{dx} = 15x^2. \]

3. Factor out \( \frac{dy}{dx} \):

   \[ \frac{dy}{dx} (2y - 3) = 15x^2. \]

4. Solve for \( \frac{dy}{dx} \):

   \[ \frac{dy}{dx} = \frac{15x^2}{2y - 3}. \]

This process allows us to find \( \frac{dy}{dx} \) even when \( y \) is not isolated. Using implicit differentiation is particularly useful for equations like circles, ellipses, and other relationships where \( y \) is not easily solved in terms of \( x \).

3.3 Differentiating Inverse Functions

To differentiate an inverse function, we use a special formula for finding the derivative of the inverse of a function. This formula is helpful when we know the function \( f(x) \), but don’t know the explicit form of its inverse \( f^{-1}(x) \).

The Derivative of an Inverse Function Formula

The formula for the derivative of an inverse function is:

\[ \frac{d}{dx} \left[ f^{-1}(x) \right] = \frac{1}{f'\left( f^{-1}(x) \right)} \]

This formula tells us that to find the derivative of \( f^{-1}(x) \) at a point, we need:

1. The derivative of \( f(x) \), denoted \( f'(x) \).
2. The value of \( f^{-1}(x) \), which is the input for \( f' \) in the denominator.

Example Problem Explanation

Let’s apply this formula to the example given in the image.

Problem Setup

The function is \( f(x) = 3x^3 + 1 \), and we are asked to find \( g'(25) \) where \( g(x) = f^{-1}(x) \) is the inverse function of \( f(x) \).

We need to find \( g'(25) \), which is the derivative of the inverse function \( g \) at \( x = 25 \).

Solution Steps

1. Use the Inverse Derivative Formula:

   According to the formula, \[ g'(x) = \frac{1}{f'\left(f^{-1}(x)\right)} \] Specifically, we need to find \( g'(25) \), so: \[ g'(25) = \frac{1}{f'\left(f^{-1}(25)\right)} \]

2. Find \( f^{-1}(25) \):

   We need to find the value of \( f^{-1}(25) \), which is the input value of \( x \) such that \( f(x) = 25 \).
   Set \( f(x) = 25 \): \[ 3x^3 + 1 = 25 \] Simplify this equation to solve for \( x \): \[ 3x^3 = 24 \] \[ x^3 = 8 \] \[ x = 2 \] So, \( f^{-1}(25) = 2 \).

3. Calculate \( f'(x) \):

   We now need \( f'(x) \), the derivative of \( f(x) \). For \( f(x) = 3x^3 + 1 \), the derivative is: \[ f'(x) = 9x^2 \]

4. Evaluate \( f'(2) \):

   Substitute \( x = 2 \) into \( f'(x) \): \[ f'(2) = 9 \cdot (2)^2 = 9 \cdot 4 = 36 \]

5. Calculate \( g'(25) \):

   Now we can use the inverse function derivative formula: \[ g'(25) = \frac{1}{f'(2)} = \frac{1}{36} \]

The value of \( g'(25) \) is \( \frac{1}{36} \).

3.4 Differentiating Inverse Trigonometric Functions

To differentiate inverse trigonometric functions, we use specific formulas for each function. These derivatives have unique forms that depend on the specific inverse trig function being differentiated. Let's go over each formula, the domains, and ranges for these functions, and apply these rules to solve typical problems.

Inverse Trig Derivatives

Here are the derivatives for the six common inverse trigonometric functions:

• Derivative of the Inverse Sine (Arcsine): \[ \frac{d}{dx} \left( \sin^{-1}(x) \right) = \frac{1}{\sqrt{1 - x^2}} \] Domain: \( -1 \leq x \leq 1 \)
• Derivative of the Inverse Cosine (Arccosine): \[ \frac{d}{dx} \left( \cos^{-1}(x) \right) = -\frac{1}{\sqrt{1 - x^2}} \] Domain: \( -1 \leq x \leq 1 \)
• Derivative of the Inverse Tangent (Arctangent): \[ \frac{d}{dx} \left( \tan^{-1}(x) \right) = \frac{1}{x^2 + 1} \] Domain: \( -\infty < x < \infty \)
• Derivative of the Inverse Cotangent (Arccotangent): \[ \frac{d}{dx} \left( \cot^{-1}(x) \right) = -\frac{1}{x^2 + 1} \] Domain: \( -\infty < x < \infty \)
• Derivative of the Inverse Secant (Arcsecant): \[ \frac{d}{dx} \left( \sec^{-1}(x) \right) = \frac{1}{|x|\sqrt{x^2 - 1}} \] Domain: \( |x| \geq 1 \)
• Derivative of the Inverse Cosecant (Arccosecant): \[ \frac{d}{dx} \left( \csc^{-1}(x) \right) = -\frac{1}{|x|\sqrt{x^2 - 1}} \] Domain: \( |x| \geq 1 \)

These formulas provide us with the instantaneous rate of change (slope) for each inverse trigonometric function. Note that the derivatives of inverse sine, cosine, secant, and cosecant include square roots, which restrict their domains to keep the values within real numbers.

Domain and Range of Inverse Trigonometric Functions

Understanding the domain and range of each inverse trig function is crucial for applying these derivatives:

• Arcsine \( y = \sin^{-1}(x) \): - Domain: \( -1 \leq x \leq 1 \) - Range: \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
• Arccosine \( y = \cos^{-1}(x) \): - Domain: \( -1 \leq x \leq 1 \) - Range: \( 0 \leq y \leq \pi \)
• Arctangent \( y = \tan^{-1}(x) \): - Domain: \( -\infty < x < \infty \) - Range: \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)
• Arccotangent \( y = \cot^{-1}(x) \): - Domain: \( -\infty < x < \infty \) - Range: \( 0 < y < \pi \)
• Arcsecant \( y = \sec^{-1}(x) \): - Domain: \( |x| \geq 1 \) - Range: \( 0 \leq y \leq \pi \), excluding \( y = \frac{\pi}{2} \)
• Arccosecant \( y = \csc^{-1}(x) \): - Domain: \( |x| \geq 1 \) - Range: \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \), excluding \( y = 0 \)

3.6 Calculating Higher-Order Derivatives

Higher-order derivatives provide insights into how a function's rate of change behaves over successive intervals. They are derivatives taken multiple times, each one describing the rate of change of the previous derivative. Higher-order derivatives are essential in various fields, from physics and engineering to mathematics and economics, as they allow us to explore concepts like acceleration, concavity, and jerk in depth.

Notation for Higher-Order Derivatives

There are several notations for expressing derivatives, as shown in the table. Here’s a breakdown of common notations for different orders of derivatives:

1. First Derivative: Represents the slope or instantaneous rate of change.
   • Notation: \( y' \), \( f'(x) \), or \( \frac{dy}{dx} \)
2. Second Derivative: Represents the rate of change of the first derivative, often related to concavity and acceleration.
   • Notation: \( y'' \), \( f''(x) \), or \( \frac{d^2y}{dx^2} \)
3. Third Derivative: Describes the rate of change of the second derivative, known as "jerk" in physics.
   • Notation: \( y''' \), \( f'''(x) \), or \( \frac{d^3y}{dx^3} \)
4. \( n \)-th Derivative:> For general higher-order derivatives, the notation \( y^{(n)} \), \( f^{(n)}(x) \), or \( \frac{d^n y}{dx^n} \) is used, where \( n \) indicates the order of the derivative.

Solution to Find \( \frac{d^2 y}{dx^2} \) for \( \sin y = x + y \)

Step 1: Differentiate Both Sides with Respect to \(x\) (First Derivative)

Starting with the equation:

\[ \sin y = x + y \]

Differentiate both sides with respect to \(x\):

\[ \cos y \cdot \frac{dy}{dx} = 1 + \frac{dy}{dx} \]

Let \( \frac{dy}{dx} = y' \) for simplicity. Then:

\[ \cos y \cdot y' = 1 + y' \]

Rearrange to solve for \( y' \):

\[ y' (\cos y - 1) = 1 \]

\[ y' = \frac{1}{\cos y - 1} \]

Step 2: Differentiate Again to Find \( \frac{d^2 y}{dx^2} \) (Second Derivative)

Now we find \( \frac{d^2 y}{dx^2} \) by differentiating \( y' \) with respect to \(x\):

Using the quotient rule on \( y' = \frac{1}{\cos y - 1} \):

\[ y'' = \frac{(0)(\cos y - 1) - (1)(-\sin y \cdot y')}{(\cos y - 1)^2} \]

Simplifying:

\[ y'' = \frac{\sin y \cdot y'}{(\cos y - 1)^2} \]

Substitute \( y' = \frac{1}{\cos y - 1} \):

\[ y'' = \frac{\sin y \cdot \frac{1}{\cos y - 1}}{(\cos y - 1)^2} \]

\[ y'' = \frac{\sin y}{(\cos y - 1)^3} \]

Final Answer

\[ \frac{d^2 y}{dx^2} = \frac{\sin y}{(\cos y - 1)^3} \]