Calculus II Comprehensive Study Guide: Inverse Functions, Exponentials, Logarithms, Inverse Trig, Hyperbolics, and L'Hopital's Rule

Course Information and Fundamental Calculus Principles

Calculus II office hours are scheduled on Monday, Tuesday, Thursday, and Friday from 2:30pm to 3:30pm, on Thursday from 11:00am to 12:00pm, and on Tuesday from 10:00am to 11:00am. Office hours take place on the 2nd Floor of Benner Tutoring.

The derivative of a function $f(x)$ represents the slope of the tangent line to the function at a given point, which defines the instantaneous rate of change (IROC). Mathematically, the derivative is defined using the limit of the difference quotient:

$f'(x) = \text{lim}_{h \to 0} \frac{f(x+h) - f(x)}{(x+h) - x} = \text{lim}_{h \to 0} \frac{f(x+h) - f(x)}{h}$

The definite integral of a function $f(x)$ over an interval $[a, b]$ represents the net area under the curve. It is defined formally as the limit of a Riemann sum as the number of subintervals $n$ approaches infinity:

$\text{int}_a^b f(x)\,dx = \text{lim}_{n \to \text{infty}} \text{sum}_{i=1}^n f(x_i^*)\,\Delta x$

The Fundamental Theorem of Calculus establishes the essential connection between differentiation and integration, showing that differentiation and integration are inverse operations.

Inverse Functions and One-to-One Properties

A function $f(x)$ possesses an inverse function, denoted $f^{-1}(x)$, if and only if $f(x)$ is a one-to-one (injective) function. A function is defined as one-to-one if distinct inputs always yield distinct outputs. Formally, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Graphically, a function is one-to-one if and only if its graph passes the horizontal line test, meaning every horizontal line intersects the graph at most once.

An inverse function completely undoes the operation performed by the original function. If $f(a) = b$, then $f^{-1}(b) = a$. The compositions of a function and its inverse yield the identity function:

$f(f^{-1}(x)) = x$

$f^{-1}(f(x)) = x$

For example, if $f(x) = x^3$, its inverse is $f^{-1}(x) = \sqrt[3]{x}$. Evaluating the compositions confirms this relationship:

$f(f^{-1}(x)) = f(\sqrt[3]{x}) = (\sqrt[3]{x})^3 = x$

$f^{-1}(f(x)) = f^{-1}(x^3) = \sqrt[3]{x^3} = x$

The domain of the inverse function $f^{-1}(x)$ is equal to the range of the original function $f(x)$, and the range of $f^{-1}(x)$ is equal to the domain of $f(x)$. It is important to distinguish notation: $f^{-1}(x)$ denotes the inverse function and is NOT equal to the reciprocal $\frac{1}{f(x)}$.

If a function assigns the same output to multiple distinct inputs, its inverse is not defined. For instance, if $f(0) = 0$, $f(1) = 0$, and $f(4) = 0$, then $f^{-1}(0)$ is not defined because the function fails the one-to-one condition.

To algebraically determine the inverse $f^{-1}(x)$ of a function $y = f(x)$, perform the following procedural steps:

1. Write the equation in terms of $y = f(x)$.

2. Swap the variables $x$ and $y$ to obtain $x = f(y)$.

3. Solve the resulting equation explicitly for $y$ in terms of $x$.

4. Replace $y$ with $f^{-1}(x)$.

Applying this procedure to $f(x) = 5x - 3$:

$y = 5x - 3$

$x = 5y - 3$

$x + 3 = 5y$

$y = \frac{x + 3}{5} \rightarrow f^{-1}(x) = \frac{x + 3}{5}$

Further fundamental algebraic and exponential inverse relationships include:

$f(x) = 2x \rightarrow f^{-1}(x) = \frac{x}{2}$

$f(x) = 2^x \rightarrow f^{-1}(x) = \log_2(x)$

Temperature conversion serves as a real-world application of inverse functions. The function converting Fahrenheit $F$ to Celsius $C$ is:

$C(F) = \frac{5}{9}(F - 32)$

Solving for $F$ in terms of $C$ yields the inverse function:

$\frac{9}{5}C = F - 32 \rightarrow F(C) = \frac{9}{5}C + 32$

Evaluating at $-159.67^\circ\text{F}$ gives $C = \frac{5}{9}(-159.67 - 32) = -106.48^\circ\text{C}$. Absolute zero corresponds to $-273.15^\circ\text{C}$, establishing the domain of the inverse function as $(-273.15, \text{infty})$.

For a linear relationship such as $5x + 3y = 6 - 7x$, rearranging terms yields $3y = 6 - 12x$, or $g(x) = 2 - 4x$. The derivative of this function is $g'(x) = -4$, and the derivative of its inverse function is $(g^{-1})'(x) = -\frac{1}{4}$.

Derivative of Inverse Functions (Theorem 7)

Theorem 7 provides a direct method to calculate the derivative of an inverse function without explicitly solving for the inverse function expression. If $f(x)$ is a one-to-one differentiable function with inverse $f^{-1}(x)$, then the derivative of the inverse function at a point $a$ is given by:

$(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$

provided that the denominator $f'(f^{-1}(a)) \neq 0$.

This theorem is derived directly from the limit definition of the derivative. Let (f^{-1})'(a) = \text{lim}_{x \to a} \frac{f^{-1}(x) - f^{-1}(a)}{x - a}$. Defining y = f^{-1}(x)$and$b = f^{-1}(a)$implies$f(y) = x$and$f(b) = a$. As$x \to a$,$y \to b. Substituting these terms into the limit gives:\n\n(f^{-1})'(a) = \text{lim}{y \to b} \frac{y - b}{f(y) - f(b)} = \text{lim}{y \to b} \frac{1}{\frac{f(y) - f(b)}{y - b}} = \frac{1}{f'(b)} = \frac{1}{f'(f^{-1}(a))}\n\nConsider f(x) = 2x + \cos(x)$. To find$(f^{-1})'(1)$, first evaluate$f'(x) = 2 - \sin(x)$. Because$\sin(x) \le 1$,$f'(x) \ge 1 > 0$for all$x$, meaning$f(x)$is strictly increasing and therefore one-to-one. Next, find the value$b = f^{-1}(1)$such that$f(b) = 1$. By inspection,$f(0) = 2(0) + \cos(0) = 1$, so$f^{-1}(1) = 0. Applying Theorem 7:\n\n(f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))} = \frac{1}{f'(0)} = \frac{1}{2 - \sin(0)} = \frac{1}{2}\n\nFor f(x) = 5x - 3$, evaluate$(f^{-1})'(2)$. Solving$f(x) = 2$gives$5x - 3 = 2 \rightarrow x = 1$, so$f^{-1}(2) = 1$. Since$f'(x) = 5, Theorem 7 yields:\n\n(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(1)} = \frac{1}{5}\n\nFor f(x) = x^3$at$a = 8$, the derivative is$f'(x) = 3x^2$. Since$f'(x) \ge 0$,$f(x)$is strictly increasing and one-to-one. Solving$f(x) = 8$gives$x = 2$, so$f^{-1}(8) = 2$. Evaluating$f'(2) = 3(2)^2 = 12, Theorem 7 gives:\n\n(f^{-1})'(8) = \frac{1}{f'(f^{-1}(8))} = \frac{1}{f'(2)} = \frac{1}{12}\n\nFor f(x) = \sqrt{x-2}$at$a = 2$,$f(x)$is strictly increasing for$x \ge 2$and passes the horizontal line test. Solving$f(x) = 2$gives$\sqrt{x-2} = 2 \rightarrow x - 2 = 4 \rightarrow x = 6$, so$f^{-1}(2) = 6$. The derivative is$f'(x) = \frac{1}{2\sqrt{x-2}}$, so$f'(6) = \frac{1}{2\sqrt{4}} = \frac{1}{4}. Applying Theorem 7:\n\n(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(6)} = \frac{1}{1/4} = 4\n\nAlternatively, finding the inverse explicitly gives y = \sqrt{x-2} \rightarrow x = y^2 + 2 \rightarrow f^{-1}(x) = x^2 + 2$. Differentiating gives$\frac{d}{dx}[f^{-1}(x)] = 2x$, and evaluating at$x = 2$yields$2(2) = 4, agreeing with Theorem 7.\n\nGeneral point evaluations demonstrate inverse properties directly: if f(3) = 15$, then$f^{-1}(15) = 3$. If$f^{-1}(1) = 6$, then$f(6) = 1$.

Exponential Functions and the Natural Base e

An exponential function has the general form $f(x) = a^x$, where the base $a > 0$ and $a \neq 1$. All exponential functions of the form $f(x) = a^x$ pass through the y-intercept $(0, 1)$.

The end behavior of exponential functions depends on the value of the base $a$:

If $a > 1$:

$\text{lim}_{x \to \text{infty}} a^x = \text{infty}$

$\text{lim}_{x \to -\text{infty}} a^x = 0$

If $0 < a < 1$:

$\text{lim}_{x \to \text{infty}} a^x = 0$

$\text{lim}_{x \to -\text{infty}} a^x = \text{infty}$

Attempting to calculate the derivative of $f(x) = 2^x$ using the limit definition yields:

$f'(x) = \text{lim}_{h \to 0} \frac{2^{x+h} - 2^x}{h} = \text{lim}_{h \to 0} \frac{2^x \cdot 2^h - 2^x}{h} = \text{lim}_{h \to 0} \frac{2^x(2^h - 1)}{h} = 2^x \left(\text{lim}_{h \to 0} \frac{2^h - 1}{h}\right)$

The value of the limit $\text{lim}_{h \to 0} \frac{a^h - 1}{h}$ represents the slope of the tangent line to $y = a^x$ at $x = 0$. Numerical approximations of this limit for various bases $a$ yield:

For $a = 2$: $\text{lim}_{h \to 0} \frac{2^h - 1}{h} \approx 0.6931$

For $a = 3$: $\text{lim}_{h \to 0} \frac{3^h - 1}{h} \approx 1.0936$

For $a = 10$: $\text{lim}_{h \to 0} \frac{10^h - 1}{h} \approx 2.3025$

For $a = 100$: $\text{lim}_{h \to 0} \frac{100^h - 1}{h} \approx 4.6052$

The natural exponential base $e$ is an irrational number ($e \approx 2.7182818...$) defined specifically such that this limit equals exactly $1$:

$\text{lim}_{h \to 0} \frac{e^h - 1}{h} = 1$

Consequently, the natural exponential function $f(x) = e^x$ is its own derivative:

$\frac{d}{dx}[e^x] = e^x$

Basic laws of exponents apply to exponential functions:

$e^{x+y} = e^x \cdot e^y$

$e^0 = 1$

$e^{x-y} = \frac{e^x}{e^y}$

Standard differentiation rules combined with exponential functions include:

Power Rule: $\frac{d}{dx}[x^n] = n x^{n-1}$

Sum Rule: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$; e.g., $\frac{d}{dx}[e^x + x^9] = e^x + 9x^8$

Constant Multiple Rule: $\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$; e.g., $\frac{d}{dx}[5e^x] = 5e^x$

Product Rule: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$; e.g., $\frac{d}{dx}[\sin(x)e^x] = \cos(x)e^x + \sin(x)e^x$

Quotient Rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$; e.g., $\frac{d}{dx}\left[\frac{e^x}{\cos(x)}\right] = \frac{e^x \cos(x) - e^x(-\sin(x))}{\cos^2(x)} = \frac{e^x(\cos(x) + \sin(x))}{\cos^2(x)}$

Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$; e.g., for $f(x) = e^x$ and $g(x) = 2x^2 + 1$, $\frac{d}{dx}[e^{2x^2+1}] = e^{2x^2+1} \cdot 4x = 4x e^{2x^2+1}$

Integration involving exponential functions frequently uses u-substitution. For example, to evaluate $\text{int} x e^{x^2}\,dx$, set $u = x^2$, so $du = 2x\,dx$ or $x\,dx = \frac{1}{2}\,du$:

$\text{int} x e^{x^2}\,dx = \frac{1}{2} \text{int} e^u\,du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$

To find an exponential model $f(x) = C b^x$ passing through given points $(1, 6)$ and $(3, 24)$, set up a system of equations:

$f(1) = C b^1 = 6$

$f(3) = C b^3 = 24$

Dividing the second equation by the first yields $\frac{C b^3}{C b} = \frac{24}{6} \rightarrow b^2 = 4 \rightarrow b = 2$. Substituting $b = 2$ into $C(2) = 6$ gives $C = 3$. Thus, the function is $f(x) = 3 \cdot 2^x$.

Any exponential base $a^x$ can be converted to base $e$ using the identity $a^x = e^{\text{ln}(a) \cdot x}$. Differentiating $f(x) = 2^x$ by converting base gives:

$\frac{d}{dx}[2^x] = \frac{d}{dx}\left[e^{\text{ln}(2) \cdot x}\right] = \text{ln}(2) \cdot e^{\text{ln}(2) \cdot x} = \text{ln}(2) \cdot 2^x$

In general, the derivative and integral rules for general exponential functions $a^x$ ($a > 0, a \neq 1$) are:

$\frac{d}{dx}[a^x] = \text{ln}(a) \cdot a^x$

$\text{int} a^x\,dx = \frac{a^x}{\text{ln}(a)} + C$

For example, \text{int} 2^x\,dx = \frac{2^x}{\text{ln}(2)} + C$.\n\n# Logarithmic Functions and Logarithmic Differentiation\n\nThe logarithmic function with base a$, denoted$y = \log_a(x)$, is the inverse of the exponential function$y = a^x$. Therefore,$\log_a(x) = y \iff a^y = x$. The natural logarithmic function uses base$e$and is written as$\log_e(x) = \text{ln}(x).\n\nKey properties of logarithms include:\n\n1. \log_a(a^x) = x$and$\text{ln}(e^x) = x\n\n2. a^{\log_a(x)} = x$and$e^{\text{ln}(x)} = x\n\n3. \log_a(x \cdot y) = \log_a(x) + \log_a(y)\n\n4. \log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)\n\n5. \log_a(x^r) = r \log_a(x)\n\nFor all logarithmic functions y = \log_a(x)$, the domain is$(0, \text{infty})$and the range is$(-\text{infty}, \text{infty})$. If$a > 1$, the limit behavior is$\text{lim}{x \to \text{infty}} \log_a(x) = \text{infty}$and$\text{lim}{x \to 0^+} \log_a(x) = -\text{infty}.\n\nThe Change of Base Formula allows conversion between logarithmic bases:\n\n\log_a(x) = \frac{\text{ln}(x)}{\text{ln}(a)}\n\nFor example, \log_7(20) = \frac{\text{ln}(20)}{\text{ln}(7)}.\n\nTo derive the derivative of \text{ln}(x)$, Theorem 7 or implicit differentiation can be applied. Using implicit differentiation, let$y = \text{ln}(x)$, which implies$e^y = x$. Differentiating both sides with respect to$x gives:\n\ne^y \cdot \frac{dy}{dx} = 1 \rightarrow \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\n\nThus, the derivative of the natural logarithm is:\n\n\frac{d}{dx}[\text{ln}(x)] = \frac{1}{x}\n\nCombining this with the change of base formula gives the derivative for general logarithmic functions:\n\n\frac{d}{dx}[\log_a(x)] = \frac{d}{dx}\left[\frac{\text{ln}(x)}{\text{ln}(a)}\right] = \frac{1}{x \text{ln}(a)}\n\nIncluding absolute value, \frac{d}{dx}[\text{ln}|x|] = \frac{1}{x}$for all$x \neq 0$. If$x > 0$,$\text{ln}|x| = \text{ln}(x)$, giving derivative$\frac{1}{x}$. If$x < 0$,$\text{ln}|x| = \text{ln}(-x)$, giving derivative$\frac{1}{-x} \cdot (-1) = \frac{1}{x}.\n\nThis leads directly to the integration rule for \frac{1}{x}:\n\n\text{int} \frac{1}{x}\,dx = \text{ln}|x| + C\n\nRecall that the general Power Rule for Integrals states \text{int} x^n\,dx = \frac{x^{n+1}}{n+1} + C$for all$n \neq -1$. When$n = -1$, the integral evaluates to$\text{ln}|x| + C$.

Logarithmic integration allows evaluation of trigonometric integrals such as $\text{int} \tan(x)\,dx$:

$\text{int} \tan(x)\,dx = \text{int} \frac{\sin(x)}{\cos(x)}\,dx$

Let $u = \cos(x)$, so $du = -\sin(x)\,dx$ or $\sin(x)\,dx = -du$:

$\text{int} \tan(x)\,dx = -\text{int} \frac{1}{u}\,du = -\text{ln}|u| + C = -\text{ln}|\cos(x)| + C = \text{ln}|\sec(x)| + C$

Logarithmic Differentiation is a technique used to differentiate complex products, quotients, and powers. The procedure consists of three steps:

1. Take the natural logarithm $\text{ln}$ of both sides of $y = f(x)$ and use logarithmic properties to expand the right side.

2. Differentiate implicitly with respect to $x$.

3. Solve for $\frac{dy}{dx}$ and substitute $y = f(x)$ back into the final expression.

To differentiate $y = \frac{\sin^2(x) \tan^4(x)}{(x^2+1)^2}$ using logarithmic differentiation:

$\text{ln}(y) = \text{ln}\left(\frac{\sin^2(x) \tan^4(x)}{(x^2+1)^2}\right)$

$\text{ln}(y) = 2 \text{ln}(\sin(x)) + 4 \text{ln}(\tan(x)) - 2 \text{ln}(x^2+1)$

Differentiating implicitly:

$\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{\cos(x)}{\sin(x)} + 4 \cdot \frac{\sec^2(x)}{\tan(x)} - 2 \cdot \frac{2x}{x^2+1}$

$\frac{1}{y} \frac{dy}{dx} = 2 \cot(x) + 4 \frac{\sec^2(x)}{\tan(x)} - \frac{4x}{x^2+1}$

$\frac{dy}{dx} = \left(\frac{\sin^2(x) \tan^4(x)}{(x^2+1)^2}\right) \left[2 \cot(x) + 4 \frac{\sec^2(x)}{\tan(x)} - \frac{4x}{x^2+1}\right]$

Logarithmic equations and critical numbers can be solved using logarithmic transformations. To find critical numbers of $f(x) = (5x^2 - 7x)e^x$, set $f'(x) = 0$:

$f'(x) = (10x - 7)e^x + (5x^2 - 7x)e^x = e^x(5x^2 + 3x - 7)$

Since $e^x \neq 0$, critical numbers occur where $5x^2 + 3x - 7 = 0$. Applying the quadratic formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(5)(-7)}}{2(5)} = \frac{-3 \pm \sqrt{149}}{10}$

To solve $\text{ln}(x) + \text{ln}(x-1) = 1$, combine terms using log properties:

$\text{ln}(x(x-1)) = 1 \rightarrow x(x-1) = e^1 \rightarrow x^2 - x - e = 0$

Applying the quadratic formula gives $x = \frac{1 \pm \sqrt{1 + 4e}}{2}$. Because $\text{ln}(x)$ requires $x > 0$ and $x - 1 > 0$, the only valid domain solution is $x = \frac{1 + \sqrt{1 + 4e}}{2}$.

To solve $e^{3x-1} = 10$, take $\text{ln}$ of both sides:

$3x - 1 = \text{ln}(10) \rightarrow 3x = \text{ln}(10) + 1 \rightarrow x = \frac{\text{ln}(10) + 1}{3}$

To find the inverse of $f(x) = \sqrt{3 - e^{2x}}$, set $y = \sqrt{3 - e^{2x}}$, square both sides to get $y^2 = 3 - e^{2x}$, rearrange to $e^{2x} = 3 - y^2$, take $\text{ln}$ to get $2x = \text{ln}(3 - y^2)$, and swap variables to obtain $f^{-1}(x) = \frac{1}{2}\text{ln}(3 - x^2)$.

Inverse Trigonometric Functions

The inverse trigonometric functions restrict domains to ensure one-to-one behavior. Derivatives of inverse trigonometric functions are derived using implicit differentiation.

To derive $\frac{d}{dx}[\csc^{-1}(x)]$, let $y = \csc^{-1}(x)$, so $\csc(y) = x$. Differentiating implicitly gives:

$-\csc(y)\cot(y) \frac{dy}{dx} = 1 \rightarrow \frac{dy}{dx} = -\frac{1}{\csc(y)\cot(y)}$

Using the Pythagorean identity $\cot^2(y) + 1 = \csc^2(y)$, $\cot(y) = \sqrt{x^2 - 1}$. Substituting gives:

$\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{|x|\sqrt{x^2 - 1}}$

The primary required inverse trigonometric derivatives are:

$\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}$

To find the derivative of $y = \sin^{-1}(2x+1)$, apply the chain rule:

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

To find the derivative of $y = x \tan^{-1}(x)$, apply the product rule:

$\frac{dy}{dx} = (1)\tan^{-1}(x) + x \cdot \frac{1}{1+x^2} = \tan^{-1}(x) + \frac{x}{1+x^2}$

Integration formulas giving inverse trigonometric functions include:

$\text{int} \frac{1}{1+4x^2}\,dx$

Let $u = 2x$, so $du = 2\,dx$ or $dx = \frac{1}{2}\,du$:

$\text{int} \frac{1}{1+4x^2}\,dx = \frac{1}{2} \text{int} \frac{1}{1+u^2}\,du = \frac{1}{2} \tan^{-1}(u) + C = \frac{1}{2} \tan^{-1}(2x) + C$

In contrast, $\text{int} \frac{x}{1+4x^2}\,dx$ uses $u = 1+4x^2$, so $du = 8x\,dx$:

$\text{int} \frac{x}{1+4x^2}\,dx = \frac{1}{8} \text{int} \frac{1}{u}\,du = \frac{1}{8} \text{ln}|u| + C = \frac{1}{8} \text{ln}(1+4x^2) + C$

The general integration formula for inverse tangent is:

$\text{int} \frac{1}{a^2 + x^2}\,dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$

Limits involving inverse trigonometric functions rely on asymptotic behavior. For $\text{lim}_{z \to \text{infty}} \tan^{-1}(\text{ln}(z))$, as $z \to \text{infty}$, $\text{ln}(z) \to \text{infty}$. As $t \to \text{infty}$, $\tan^{-1}(t) \to \frac{\pi}{2}$. Therefore:

$\text{lim}_{z \to \text{infty}} \tan^{-1}(\text{ln}(z)) = \frac{\pi}{2}$

Hyperbolic Functions

Hyperbolic functions are defined using linear combinations of exponential functions $e^x$ and $e^{-x}$:

Hyperbolic Sine ("Sinh"): $\sinh(x) = \frac{e^x - e^{-x}}{2}$

Hyperbolic Cosine ("Cosh"): $\cosh(x) = \frac{e^x + e^{-x}}{2}$

Hyperbolic Tangent ("Tanh"): $\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

While trigonometric functions parameterize a unit circle ($x^2 + y^2 = 1$ via $x = \cos(t), y = \sin(t)$), hyperbolic functions parameterize a unit hyperbola ($x^2 - y^2 = 1$ via $x = \cosh(t), y = \sinh(t)$). The fundamental hyperbolic identity is:

$\cosh^2(x) - \sinh^2(x) = 1$

This identity is verified algebraically:

$\left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1$

Symmetry properties show that $\sinh(x)$ is an odd function and $\cosh(x)$ is an even function:

$\sinh(-x) = \frac{e^{-x} - e^x}{2} = -\sinh(x)$

$\cosh(-x) = \frac{e^{-x} + e^x}{2} = \cosh(x)$

Derivatives of hyperbolic functions are calculated directly from their exponential definitions:

$\frac{d}{dx}[\sinh(x)] = \frac{d}{dx}\left[\frac{e^x - e^{-x}}{2}\right] = \frac{e^x + e^{-x}}{2} = \cosh(x)$

$\frac{d}{dx}[\cosh(x)] = \frac{d}{dx}\left[\frac{e^x + e^{-x}}{2}\right] = \frac{e^x - e^{-x}}{2} = \sinh(x)$

$\frac{d}{dx}[\tanh(x)] = \frac{d}{dx}\left[\frac{\sinh(x)}{\cosh(x)}\right] = \frac{\cosh(x)\cosh(x) - \sinh(x)\sinh(x)}{\cosh^2(x)} = \frac{\cosh^2(x) - \sinh^2(x)}{\cosh^2(x)} = \frac{1}{\cosh^2(x)} = \text{sech}^2(x)$

Inverse hyperbolic functions can be expressed explicitly as logarithmic functions. The formula for inverse hyperbolic sine is:

$\sinh^{-1}(x) = \text{ln}\left(x + \sqrt{x^2+1}\right)$

To differentiate $y = \text{ln}(\cosh(x))$, use the chain rule:

$\frac{dy}{dx} = \frac{1}{\cosh(x)} \cdot \sinh(x) = \tanh(x)$

To differentiate $y = x \sinh^{-1}\left(\frac{x}{3}\right) - \sqrt{9+x^2}$:

$\frac{dy}{dx} = (1)\sinh^{-1}\left(\frac{x}{3}\right) + x \cdot \frac{1/3}{\sqrt{1 + (x/3)^2}} - \frac{2x}{2\sqrt{9+x^2}} = \sinh^{-1}\left(\frac{x}{3}\right) + \frac{x}{\sqrt{9+x^2}} - \frac{x}{\sqrt{9+x^2}} = \sinh^{-1}\left(\frac{x}{3}\right)$

Definite integration with hyperbolic inverse functions yields exact values:

$\text{int}_0^1 \frac{1}{\sqrt{1+x^2}}\,dx = \left[\sinh^{-1}(x)\right]_0^1 = \left[\text{ln}\left(x + \sqrt{x^2+1}\right)\right]_0^1 = \text{ln}(1+\sqrt{2}) - \text{ln}(1) = \text{ln}(1+\sqrt{2})$

For integrals of the form $\text{int} \frac{e^x}{1 - e^{2x}}\,dx$, substitute $u = e^x, du = e^x\,dx$:

$\text{int} \frac{1}{1-u^2}\,du = \tanh^{-1}(u) + C = \tanh^{-1}(e^x) + C$

Physical applications of hyperbolic functions include describing catenary curves formed by hanging cables, modeled by equations such as f(x) = 20 \cosh\left(\frac{x}{20}\right) - 15$.\n\n# Indeterminate Forms and L'Hôpital's Rule\n\nIf \text{lim}{x \to a} f(x) = 0$and$\text{lim}{x \to a} g(x) = 0$, then$\text{lim}{x \to a} \frac{f(x)}{g(x)}$is an indeterminate form of type$\frac{0}{0}$. If$\text{lim}{x \to a} f(x) = \pm\text{infty}$and$\text{lim}{x \to a} g(x) = \pm\text{infty}$, then$\text{lim}{x \to a} \frac{f(x)}{g(x)}$is an indeterminate form of type$\frac{\text{infty}}{\text{infty}}.\n\nL'Hôpital's Rule states that if f$and$g$are differentiable functions and$\text{lim}{x \to a} \frac{f(x)}{g(x)}$is of type$\frac{0}{0}$or$\frac{\text{infty}}{\text{infty}}, then:\n\n\text{lim}{x \to a} \frac{f(x)}{g(x)} = \text{lim}{x \to a} \frac{f'(x)}{g'(x)}\n\nprovided the limit on the right exists or is infinite.\n\nThe proof of L'Hôpital's Rule relies on Cauchy's Mean Value Theorem. Assuming f(a) = 0$and$g(a) = 0$, Cauchy's MVT states that there exists a number$y$between$a$and$x such that:\n\n\frac{f'(y)}{g'(y)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f(x)}{g(x)}\n\nTaking the limit as x \to a$causes$y \to a$, proving that$\text{lim}{x \to a} \frac{f(x)}{g(x)} = \text{lim}{x \to a} \frac{f'(x)}{g'(x)}.\n\nExamples of basic indeterminate forms include:\n\n\text{lim}{x \to 0} \frac{\text{ln}(x+1)}{x} \stackrel{L'H}{=} \text{lim}{x \to 0} \frac{\frac{1}{x+1}}{1} = 1\n\n\text{lim}{x \to 0} \frac{\sin(x)}{x} \stackrel{L'H}{=} \text{lim}{x \to 0} \frac{\cos(x)}{1} = 1\n\n\text{lim}{t \to 0} \frac{e^t - t - 1}{t^2} \stackrel{L'H}{=} \text{lim}{t \to 0} \frac{e^t - 1}{2t} \stackrel{L'H}{=} \text{lim}{t \to 0} \frac{e^t}{2} = \frac{1}{2}\n\n\text{lim}{t \to \text{infty}} \frac{e^t}{t^2} \stackrel{L'H}{=} \text{lim}{t \to \text{infty}} \frac{e^t}{2t} \stackrel{L'H}{=} \text{lim}{t \to \text{infty}} \frac{e^t}{2} = \text{infty}\n\nIndeterminate Products (0 \cdot \text{infty}$) are rewritten as quotients$f(x)g(x) = \frac{f(x)}{1/g(x)}$to create type$\frac{0}{0}$or$\frac{\text{infty}}{\text{infty}}.\n\n\text{lim}{x \to \text{infty}} x e^{-x} = \text{lim}{x \to \text{infty}} \frac{x}{e^x} \stackrel{L'H}{=} \text{lim}{x \to \text{infty}} \frac{1}{e^x} = 0\n\nIndeterminate Differences (\text{infty} - \text{infty}) are resolved by finding a common denominator to unify the expression into a single quotient.\n\n\text{lim}{x \to 1} \left(\frac{x}{x-1} - \frac{1}{\text{ln}(x)}\right) = \text{lim}{x \to 1} \frac{x\text{ln}(x) - (x-1)}{(x-1)\text{ln}(x)}\n\nApplying L'Hôpital's Rule iteratively:\n\n\stackrel{L'H}{=} \text{lim}{x \to 1} \frac{\text{ln}(x) + 1 - 1}{\text{ln}(x) + (x-1)\frac{1}{x}} = \text{lim}{x \to 1} \frac{\text{ln}(x)}{\text{ln}(x) + 1 - \frac{1}{x}} \stackrel{L'H}{=} \text{lim}_{x \to 1} \frac{\frac{1}{x}}{\frac{1}{x} + \frac{1}{x^2}} = \frac{1}{1 + 1} = \frac{1}{2}\n\nIndeterminate Powers (0^0, \text{infty}^0, 1^\text{infty}$) are solved by taking the natural logarithm of$y = f(x)^{g(x)}$, converting the power into a product$\text{ln}(y) = g(x)\text{ln}(f(x))$, evaluating the limit$L = \text{lim} \text{ln}(y)$, and exponentiating the result$y = e^L$.

For $\text{lim}_{x \to 0^+} x^x$ (Type $0^0$):

Let $y = x^x \rightarrow \text{ln}(y) = x \text{ln}(x)$

$\text{lim}_{x \to 0^+} \text{ln}(y) = \text{lim}_{x \to 0^+} \frac{\text{ln}(x)}{\frac{1}{x}} \stackrel{L'H}{=} \text{lim}_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \text{lim}_{x \to 0^+} (-x) = 0$

Since $\text{ln}(y) \to 0$, $y \to e^0 = 1$. Therefore, \text{lim}_{x \to 0^+} x^x = 1$.\n\nFor \text{lim}{x \to 0} (1 - 2x)^{1/x}$(Type$1^\text{infty}):\n\nLet y = (1 - 2x)^{1/x} \rightarrow \text{ln}(y) = \frac{\text{ln}(1-2x)}{x}\n\n\text{lim}{x \to 0} \text{ln}(y) = \text{lim}{x \to 0} \frac{\text{ln}(1-2x)}{x} \stackrel{L'H}{=} \text{lim}{x \to 0} \frac{\frac{-2}{1-2x}}{1} = -2\n\nSince \text{ln}(y) \to -2$,$y \to e^{-2} = \frac{1}{e^2}.\n\n# Practice Problems and Solutions\n\nProblem 1: Determine if f(x) = 3 - 2x is one-to-one. If so, find its inverse.\n\nSolution: The linear function f(x) = 3 - 2x$has a constant negative slope ($-2$), making it strictly decreasing across its entire domain. It passes the horizontal line test and is one-to-one. Set$y = 3 - 2x$, swap variables to$x = 3 - 2y$, and solve for$y:\n\n2y = 3 - x \rightarrow y = -\frac{1}{2}x + \frac{3}{2}\n\nf^{-1}(x) = -\frac{1}{2}x + \frac{3}{2}\n\nProblem 2: Evaluate \cos^{-1}(-1).\n\nSolution: The range of \cos^{-1}(x)$is restricted to$[0, \pi]$. The angle$\theta \in [0, \pi]$satisfying$\cos(\theta) = -1$is$\pi$. Therefore,$\cos^{-1}(-1) = \pi$.

Problem 3: Solve for $x$ in the equation \frac{10}{1 + e^{-x}} = 3$.\n\nSolution: Multiply both sides by (1 + e^{-x}) and isolate the exponential term:\n\n10 = 3(1 + e^{-x}) \rightarrow 10 = 3 + 3e^{-x} \rightarrow 7 = 3e^{-x} \rightarrow e^{-x} = \frac{7}{3}\n\nTake the natural logarithm of both sides:\n\n-x = \text{ln}\left(\frac{7}{3}\right) \rightarrow x = -\text{ln}\left(\frac{7}{3}\right) = \text{ln}\left(\frac{3}{7}\right)\n\nProblem 4: Find the derivative of f(t) = \frac{1 + e^t}{1 - e^t}$.

Solution: Apply the Quotient Rule:

$f'(t) = \frac{e^t(1 - e^t) - (1 + e^t)(-e^t)}{(1 - e^t)^2} = \frac{e^t - e^{2t} + e^t + e^{2t}}{(1 - e^t)^2} = \frac{2e^t}{(1 - e^t)^2}$