Intermediate Value Theorem, Higher Derivatives, and the Chain Rule in Calculus

Lecture Learning Outcomes

• Apply the Intermediate Value Theorem

• Define the Chain Rule and apply it to find derivatives of composite functions

The Intermediate Value Theorem

Statement and Explanation

The Intermediate Value Theorem (IVT) is a fundamental result in calculus concerning continuous functions. It states:

• If f is continuous on the closed interval [a, b], and N is any number between f(a) and f(b) (where f(a) ≠ f(b)), then there exists at least one number c in (a, b) such that f(c) = N.

This theorem guarantees that a continuous function attains every value between f(a) and f(b) at some point in the interval.

• Continuity is essential for the theorem to hold.

• Commonly used to prove the existence of roots (solutions) to equations.

Graphical Illustration

• Consider a function y = f(x) that is continuous on [a, b].

• If f(a) < N < f(b), then there is some c in (a, b) where f(c) = N.

Example: Existence of a Root

Use the Intermediate Value Theorem to show that there is a solution to -x^3 + 4x + 1 = 0 in the interval (-1, 0).

• Let f(x) = -x^3 + 4x + 1. This is a polynomial, so it is continuous everywhere.

• Evaluate at endpoints:

  • f(-1) = -(-1)^3 + 4(-1) + 1 = 1 - 4 + 1 = -2

  • f(0) = -(0)^3 + 4(0) + 1 = 1

• Since f(-1) = -2 < 0 < 1 = f(0), by IVT, there exists c in (-1, 0) such that f(c) = 0.

Higher Derivatives

Definition and Notation

The derivative of a function measures its rate of change. Higher derivatives are obtained by differentiating a function multiple times.

• First derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

• Second derivative: $f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h}$

• Third derivative: $f'''(x) = \lim_{h \to 0} \frac{f''(x+h) - f''(x)}{h}$

Higher derivatives are used to analyze concavity, inflection points, and motion in physics.

Example: Compute Higher Derivatives

Given f(x) = 4x^6 - 6\sqrt{x}, find f'(x), f''(x), and f'''(x):

• $f'(x) = 24x^5 - 3x^{-1/2}$

• $f''(x) = 120x^4 + \frac{3}{2}x^{-3/2}$

• $f'''(x) = 480x^3 - \frac{9}{4}x^{-5/2}$

The Chain Rule

Statement and Explanation

The Chain Rule is a formula for computing the derivative of a composite function. If y = f(g(x)), then:

• $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

• In Leibniz notation, if y = f(u) and u = g(x):

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

This rule is essential for differentiating functions within functions, such as trigonometric, exponential, and logarithmic compositions.

Examples of the Chain Rule

• Example 1: $y = (x^2 + 1)^2$

  • Let $u = x^2 + 1$, then $y = u^2$

  • $\frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)$

• Example 2: $y = \sin(x^2)$

  • $\frac{dy}{dx} = \cos(x^2) \cdot 2x$

• Example 3: $y = e^{\sin x}$

  • $\frac{dy}{dx} = e^{\sin x} \cdot \cos x$

• Example 4: $y = \tan(2x)$

  • $\frac{dy}{dx} = \sec^2(2x) \cdot 2$

Further Examples

• Differentiate $y = (\sin x)^2$:

  • $\frac{dy}{dx} = 2\sin x \cdot \cos x$

• Differentiate $y = \sqrt{x^4 + 2x}$:

  • $\frac{dy}{dx} = \frac{1}{2}(x^4 + 2x)^{-1/2} \cdot (4x^3 + 2)$

• Differentiate $U(y) = \left(\frac{y^4 + 1}{y^2 + 1}\right)^5$:

  • $U'(y) = 5\left(\frac{y^4 + 1}{y^2 + 1}\right)^4 \cdot \frac{(y^2 + 1)(4y^3) - (y^4 + 1)(2y)}{(y^2 + 1)^2}$

• Differentiate $y = (2x + 1)^5 (x^3 - x + 1)^4$:

  • Apply the product and chain rules:

  • $y' = (2x + 1)^5 \cdot 4(x^3 - x + 1)^3 \cdot (3x^2 - 1) + (x^3 - x + 1)^4 \cdot 5(2x + 1)^4 \cdot 2$

Summary Table: Chain Rule Applications

Function Type	Derivative	Example
Power of a function	$y = [f(x)]^n \implies y' = n[f(x)]^{n-1} f'(x)$	$y = (x^2 + 1)^3 \implies y' = 3(x^2 + 1)^2 \cdot 2x$
Exponential of a function	$y = e^{f(x)} \implies y' = e^{f(x)} f'(x)$	$y = e^{x^2} \implies y' = e^{x^2} \cdot 2x$
Trigonometric of a function	$y = \sin(f(x)) \implies y' = \cos(f(x)) f'(x)$	$y = \sin(x^2) \implies y' = \cos(x^2) \cdot 2x$

Additional info: The notes also include several worked examples and step-by-step applications of the chain rule, product rule, and higher derivatives, which are essential for mastering differentiation techniques in calculus.