Calculus I: Differentiation, Applications, and Extrema – Exam Review Study Notes

Differentiation of Common Functions

Derivatives of Trigonometric, Inverse Trigonometric, and Logarithmic Functions

The derivative rules for trigonometric, inverse trigonometric, and logarithmic functions are essential tools in calculus. Mastery of these formulas allows for efficient computation and analysis of rates of change in various mathematical contexts.

• Trigonometric Functions:

  • $\frac{d}{dx}(\sin x) = \cos x$

  • $\frac{d}{dx}(\cos x) = -\sin x$

  • $\frac{d}{dx}(\tan x) = \sec^2 x$

  • $\frac{d}{dx}(\cot x) = -\csc^2 x$

  • $\frac{d}{dx}(\sec x) = \sec x \tan x$

  • $\frac{d}{dx}(\csc x) = -\csc x \cot x$

• Inverse Trigonometric Functions:

  • $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$

  • $\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}$

  • $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$

  • $\frac{d}{dx}(\arccot x) = -\frac{1}{1+x^2}$

  • $\frac{d}{dx}(\arcsec x) = \frac{1}{|x|\sqrt{x^2-1}}$

  • $\frac{d}{dx}(\arccsc x) = -\frac{1}{|x|\sqrt{x^2-1}}$

• Logarithmic and Exponential Functions:

  • $\frac{d}{dx}(\ln x) = \frac{1}{x}$

  • $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$

  • $\frac{d}{dx}(e^x) = e^x$

  • $\frac{d}{dx}(a^x) = a^x \ln a$

Example: Find $\frac{d}{dx}(\arctan x^2)$. Let $u = x^2$, then $\frac{d}{dx}(\arctan u) = \frac{1}{1+u^2} \cdot \frac{du}{dx} = \frac{1}{1+x^4} \cdot 2x = \frac{2x}{1+x^4}$.

Logarithmic Differentiation

Definition and Application

Logarithmic differentiation is a technique for differentiating functions by first taking the natural logarithm of both sides, then differentiating implicitly. This is especially useful for functions involving products, quotients, or powers.

• General Steps:

  1. Take the natural logarithm of both sides: $y = f(x) \implies \ln y = \ln f(x)$

  2. Differentiate both sides with respect to $x$ using implicit differentiation.

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

• Example: Differentiate $y = x^x$.

  • Take $\ln$ of both sides: $\ln y = x \ln x$

  • Differentiate: $\frac{1}{y} \frac{dy}{dx} = \ln x + 1$

  • So $\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

Related Rates Problems

Solving Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to another, often using implicit differentiation.

• General Approach:

  1. Identify all variables and their relationships.

  2. Differentiate both sides of the equation with respect to time $t$.

  3. Substitute known values and solve for the desired rate.

• Example: If $x^2 + y^2 = r^2$ and $x$ and $y$ are changing with time, then $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.

Increasing and Decreasing Functions

First Derivative Test

The First Derivative Test is used to determine where a function is increasing or decreasing and to identify local extrema.

• If $f'(x) > 0$ on an interval, $f(x)$ is increasing there.

• If $f'(x) < 0$ on an interval, $f(x)$ is decreasing there.

• Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined.

• Local Maximum: If $f'(x)$ changes from positive to negative at $c$, $f(c)$ is a local maximum.

• Local Minimum: If $f'(x)$ changes from negative to positive at $c$, $f(c)$ is a local minimum.

Absolute Extrema on a Closed Interval

Closed Interval Method

To find the absolute maximum and minimum values of a continuous function $f$ on a closed interval $[a, b]$, follow these steps:

1. Find the critical numbers of $f$ in $[a, b]$ (where $f'(x) = 0$ or $f'(x)$ is undefined).

2. Evaluate $f$ at each critical number and at the endpoints $a$ and $b$.

3. The largest value is the absolute maximum; the smallest is the absolute minimum.

Example: For $f(x) = x^3 - 3x^2 + 1$ on $[0, 3]$, find $f'(x) = 3x^2 - 6x$. Set $f'(x) = 0$ to find critical points, then evaluate $f$ at those points and at $x=0$ and $x=3$.

Rolle's Theorem and the Mean Value Theorem

Rolle's Theorem

Rolle's Theorem states that if $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists $c$ in $(a, b)$ such that $f'(c) = 0$.

Mean Value Theorem (MVT)

The Mean Value Theorem states that if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists $c$ in $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Application: The MVT guarantees the existence of a point where the instantaneous rate of change equals the average rate of change over the interval.

Concavity and Inflection Points

Second Derivative Test

The Second Derivative Test helps determine the concavity of a function and identify inflection points.

• If $f''(x) > 0$ on an interval, $f$ is concave up there.

• If $f''(x) < 0$ on an interval, $f$ is concave down there.

• An inflection point occurs where $f''(x)$ changes sign.

• Local Extrema: If $f'(c) = 0$ and $f''(c) > 0$, $f(c)$ is a local minimum; if $f''(c) < 0$, $f(c)$ is a local maximum.

Summary Table: Derivatives of Common Functions

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$
$\arctan x$	$\frac{1}{1+x^2}$
$\arccot x$	$-\frac{1}{1+x^2}$
$\arcsec x$	$\frac{1}{|x|\sqrt{x^2-1}}$
$\arccsc x$	$-\frac{1}{|x|\sqrt{x^2-1}}$
$\ln x$	$\frac{1}{x}$
$\log_a x$	$\frac{1}{x \ln a}$
$e^x$	$e^x$
$a^x$	$a^x \ln a$

Additional info: These notes expand on the original brief points by providing definitions, step-by-step procedures, and examples for each major topic, ensuring a comprehensive review for exam preparation.