Chapter 6: Inverse Functions - Exponential, Logarithmic, and Inverse Trigonometric Functions

Section 6.1: Inverse Functions and their Derivatives

  • Definition of a One-to-One Function: A function ff is called a one-to-one function if it never takes the same value twice; that is, f(x1)≠f(x2)f(x_1) \neq f(x_2) whenever x1≠x2x_1 \neq x_2.
  • Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than once.
  • Inverse Functions Definition: Let ff be a one-to-one function with domain AA and range BB. Then its inverse function f−1f^{-1} has domain BB and range AA and is defined by f−1(y)=x  ⟺  f(x)=yf^{-1}(y) = x \iff f(x) = y for any yy in BB.     - Domain and Range Relationship:         - Domain of f−1=Range of ff^{-1} = \text{Range of } f         - Range of f−1=Domain of ff^{-1} = \text{Domain of } f
  • Properties of Inverse Functions:     - Cancellation Equations:         - f−1(f(x))=xf^{-1}(f(x)) = x for every xx in AA.         - f(f−1(x))=xf(f^{-1}(x)) = x for every xx in BB.     - Graphs: The graph of f−1f^{-1} is obtained by reflecting the graph of ff about the line y=xy = x.
  • How to Find the Inverse Function of a One-to-One Function:     - Step 1: Write y=f(x)y = f(x).     - Step 2: Solve this equation for xx in terms of yy (if possible).     - Step 3: To express f−1f^{-1} as a function of xx, interchange xx and yy. The resulting equation is y=f−1(x)y = f^{-1}(x).     - Example: Find the inverse of y=x3+2y = x^3 + 2.         - x=y3+2x = y^3 + 2         - x−2=y3x - 2 = y^3         - y=x−23y = \sqrt[3]{x - 2}         - Hence, f−1(x)=(x−2)13f^{-1}(x) = (x - 2)^{\frac{1}{3}}.
  • Derivatives of Inverse Functions Theorem: If ff is a one-to-one differentiable function with inverse function f−1f^{-1} and f′(f−1(a))≠0f'(f^{-1}(a)) \neq 0, then the inverse function is differentiable at aa and is given by:     - (f−1)′(a)=1f′(f−1(a))(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}
  • Example of Derivative Calculation:     - Given f(y)=2y+cos⁡(y)f(y) = 2y + \cos(y), find (f−1)′(1)(f^{-1})'(1).     - Step 1: Find yy such that f(y)=1f(y) = 1. By inspection, if y=0y = 0, then 2(0)+cos⁡(0)=0+1=12(0) + \cos(0) = 0 + 1 = 1. Thus, f−1(1)=0f^{-1}(1) = 0.     - Step 2: Find f′(y)=ddy(2y+cos⁡(y))=2−sin⁡(y)f'(y) = \frac{d}{dy}(2y + \cos(y)) = 2 - \sin(y).     - Step 3: Use the formula: (f−1)′(1)=1f′(f−1(1))=1f′(0)=12−sin⁡(0)=12(f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))} = \frac{1}{f'(0)} = \frac{1}{2 - \sin(0)} = \frac{1}{2}.

Section 6.2: The Natural Logarithmic Function

  • Definition: The natural logarithmic function is the function defined by ln⁡(x)=∫1x1t dt\ln(x) = \int_{1}^{x} \frac{1}{t} \, dt for x>0x > 0.
  • Laws of Logarithms: If xx and yy are positive numbers and rr is a rational number, then:     - ln⁡(xy)=ln⁡(x)+ln⁡(y)\ln(xy) = \ln(x) + \ln(y)     - ln⁡(xy)=ln⁡(x)−ln⁡(y)\ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y)     - ln⁡(xr)=rln⁡(x)\ln(x^r) = r \ln(x)     - Note: Terms like ln⁡(x)\ln(x) or ln⁡(y)\ln(y) are undefined if x≤0x \leq 0 or y≤0y \leq 0.
  • Derivative of the Natural Logarithm:     - ddx(ln⁡(x))=1x\frac{d}{dx}(\ln(x)) = \frac{1}{x}     - General Form (Chain Rule): ddx(ln⁡(g(x)))=g′(x)g(x)\frac{d}{dx}(\ln(g(x))) = \frac{g'(x)}{g(x)}
  • Example Derivatives:     - Find ddxln⁡(x+1x−2)\frac{d}{dx} \ln\left( \frac{x+1}{\sqrt{x-2}} \right).     - Simplified using laws: ln⁡(x+1)−ln⁡((x−2)12)=ln⁡(x+1)−12ln⁡(x−2)\ln(x+1) - \ln((x-2)^{\frac{1}{2}}) = \ln(x+1) - \frac{1}{2} \ln(x-2).     - Derivative: 1x+1−12(x−2)\frac{1}{x+1} - \frac{1}{2(x-2)}.
  • Integrals involving the Natural Logarithm:     - ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln|x| + C     - ∫tan⁡(x) dx=ln⁡∣sec⁡(x)∣+C\int \tan(x) \, dx = \ln|\sec(x)| + C     - Example with Substitution: Evaluate ∫2xx2+1 dx\int \frac{2x}{x^2+1} \, dx.         - Let u=x2+1u = x^2 + 1, then du=2x dxdu = 2x \, dx.         - ∫1u du=ln⁡∣u∣+C=ln⁡(x2+1)+C\int \frac{1}{u} \, du = \ln|u| + C = \ln(x^2+1) + C (absolute value not needed as x2+1>0x^2+1 > 0).
  • Logarithmic Differentiation: A method to simplify the differentiation of products, quotients, or powers.     - Steps:         1. Take natural logarithms of both sides of y=f(x)y = f(x) and use laws of logarithms to expand the expression.         2. Differentiate implicitly with respect to xx.         3. Solve for y′y' and replace yy with f(x)f(x).     - Example: Differentiate y=x34x2+1(3x+2)5y = \frac{x^{\frac{3}{4}} \sqrt{x^2+1}}{(3x+2)^5}.         - ln⁡(y)=34ln⁡(x)+12ln⁡(x2+1)−5ln⁡(3x+2)\ln(y) = \frac{3}{4} \ln(x) + \frac{1}{2} \ln(x^2+1) - 5 \ln(3x+2).         - 1yy′=34x+xx2+1−153x+2\frac{1}{y} y' = \frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2}.         - y′=y(34x+xx2+1−153x+2)y' = y \left( \frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2} \right).

Section 6.3: The Natural Exponential Function

  • Definition: The natural exponential function f(x)=exf(x) = e^x is the inverse of the natural logarithmic function ln⁡(x)\ln(x).     - ex=y  ⟺  ln⁡(y)=xe^x = y \iff \ln(y) = x
  • Properties:     - ln⁡(ex)=x\ln(e^x) = x for x∈Rx \in \mathbb{R}     - eln⁡(x)=xe^{\ln(x)} = x for x>0x > 0     - Limit properties: lim⁡x→−∞ex=0\lim_{x \rightarrow -\infty} e^x = 0 and lim⁡x→∞ex=∞\lim_{x \rightarrow \infty} e^x = \infty
  • Differentiation:     - ddx(ex)=ex\frac{d}{dx}(e^x) = e^x     - Chain Rule: ddx(eu(x))=eu(x)⋅u′(x)\frac{d}{dx}(e^{u(x)}) = e^{u(x)} \cdot u'(x).
  • Integration:     - ∫ex dx=ex+C\int e^x \, dx = e^x + C     - Example: Evaluate ∫x2ex3 dx\int x^2 e^{x^3} \, dx.         - Let u=x3u = x^3, then du=3x2 dxdu = 3x^2 \, dx.         - 13∫eu du=13eu+C=13ex3+C\frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + C = \frac{1}{3} e^{x^3} + C.
  • Area under the Curve: Find the area under y=e−3xy = e^{-3x} from 00 to 11.     - A=∫01e−3x dx=[−13e−3x]01=(−13e−3)−(−13e0)=13(1−e−3)A = \int_{0}^{1} e^{-3x} \, dx = \left[ -\frac{1}{3} e^{-3x} \right]_{0}^{1} = \left( -\frac{1}{3} e^{-3} \right) - \left( -\frac{1}{3} e^0 \right) = \frac{1}{3} (1 - e^{-3}).

Section 6.4: General Logarithmic and Exponential Functions

  • General Exponential Function: For b>0b > 0, we define bx=exln⁡(b)b^x = e^{x \ln(b)}.
  • Differentiation of General Exponential:     - ddx(bx)=bxln⁡(b)\frac{d}{dx}(b^x) = b^x \ln(b)     - ddx(bg(x))=bg(x)ln⁡(b)⋅g′(x)\frac{d}{dx}(b^{g(x)}) = b^{g(x)} \ln(b) \cdot g'(x)
  • Exponential Graphs:     - If b>1b > 1, y=bxy = b^x is increasing. lim⁡x→−∞bx=0\lim_{x \rightarrow -\infty} b^x = 0 and lim⁡x→∞bx=∞\lim_{x \rightarrow \infty} b^x = \infty.     - If 0<b<10 < b < 1, y=bxy = b^x is decreasing. lim⁡x→−∞bx=∞\lim_{x \rightarrow -\infty} b^x = \infty and lim⁡x→∞bx=0\lim_{x \rightarrow \infty} b^x = 0.
  • General Exponential Integrals:     - ∫bx dx=bxln⁡(b)+C\int b^x \, dx = \frac{b^x}{\ln(b)} + C
  • The Power Rule versus the Exponential Rule:     1. Variable base, constant exponent: ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}.     2. Constant base, variable exponent: ddx(bx)=bxln⁡(b)\frac{d}{dx}(b^x) = b^x \ln(b).     3. Variable base, variable exponent: Use logarithmic differentiation for functions like y=(g(x))h(x)y = (g(x))^{h(x)}.         - Example: Differentiate y=(cos⁡(x))xy = (\cos(x))^x.         - ln⁡(y)=xln⁡(cos⁡(x))\ln(y) = x \ln(\cos(x))         - y′y=ln⁡(cos⁡(x))+x⋅−sin⁡(x)cos⁡(x)\frac{y'}{y} = \ln(\cos(x)) + x \cdot \frac{-\sin(x)}{\cos(x)}         - y′=(cos⁡(x))x[ln⁡(cos⁡(x))−xtan⁡(x)]y' = (\cos(x))^x [\ln(\cos(x)) - x \tan(x)]
  • General Logarithmic Function: If b>0b > 0 and b≠1b \neq 1, the inverse of f(x)=bxf(x) = b^x is y=log⁡b(x)y = \log_b(x).     - log⁡b(x)=y  ⟺  by=x\log_b(x) = y \iff b^y = x     - log⁡b(x)=ln⁡(x)ln⁡(b)\log_b(x) = \frac{\ln(x)}{\ln(b)}     - Cancellation Equations:         - blog⁡b(x)=xb^{\log_b(x)} = x         - log⁡b(bx)=x\log_b(b^x) = x
  • Derivative of Logarithm with Base bb:     - ddx(log⁡b(x))=1xln⁡(b)\frac{d}{dx}(\log_b(x)) = \frac{1}{x \ln(b)}

Section 6.8: Indeterminate Forms and l’Hospital’s Rule

  • L’Hospital’s Rule: Suppose ff and gg are differentiable and g′(x)≠0g'(x) \neq 0 near aa (except possibly at aa). If the limit lim⁡x→af(x)g(x)\lim_{x \rightarrow a} \frac{f(x)}{g(x)} results in an indeterminate form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, then:     - lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} provided the right-side limit exists.     - Note: Differentiate the numerator and denominator separately; do not use the Quotient Rule.
  • Indeterminate Forms:     - Direct Forms: 00,∞∞\frac{0}{0}, \frac{\infty}{\infty}.     - Indeterminate Product: 0⋅∞0 \cdot \infty. Rewrite as lim⁡f⋅g=lim⁡f1/g\lim f \cdot g = \lim \frac{f}{1/g} to use L’Hospital.     - Indeterminate Difference: ∞−∞\infty - \infty. Combine terms or find a common denominator.     - Indeterminate Powers: 00,∞0,1∞0^0, \infty^0, 1^\infty. Use y=f(x)g(x)y = f(x)^{g(x)}, then ln⁡(y)=g(x)ln⁡(f(x))\ln(y) = g(x) \ln(f(x)). Find the limit of ln⁡(y)\ln(y) and then find the limit of y=eln⁡(y)y = e^{\ln(y)}.
  • Example Calculations:     - Limit with ln: Evaluate lim⁡x→1ln⁡(x)x−1\lim_{x \rightarrow 1} \frac{\ln(x)}{x-1}.         - Form 0/00/0. Apply L'Hospital: lim⁡x→11/x1=1\lim_{x \rightarrow 1} \frac{1/x}{1} = 1.     - Product Example: Evaluate lim⁡x→∞xsin⁡(πx)\lim_{x \rightarrow \infty} x \sin\left(\frac{\pi}{x}\right).         - Form ∞⋅0\infty \cdot 0. Rewrite as lim⁡x→∞sin⁡(π/x)1/x\lim_{x \rightarrow \infty} \frac{\sin(\pi/x)}{1/x}.         - Let t=1/xt = 1/x, as x→∞x \rightarrow \infty, t→0+t \rightarrow 0^+.         - lim⁡t→0+sin⁡(πt)t=lim⁡t→0+πcos⁡(πt)1=π(1)=π\lim_{t \rightarrow 0^+} \frac{\sin(\pi t)}{t} = \lim_{t \rightarrow 0^+} \frac{\pi \cos(\pi t)}{1} = \pi(1) = \pi.     - Power Example: Evaluate lim⁡x→0+xx\lim_{x \rightarrow 0^+} x^x.         - Form 000^0. Let y=xxy = x^x, so ln⁡(y)=xln⁡(x)\ln(y) = x \ln(x).         - lim⁡x→0+xln⁡(x)=lim⁡x→0+ln⁡(x)1/x\lim_{x \rightarrow 0^+} x \ln(x) = \lim_{x \rightarrow 0^+} \frac{\ln(x)}{1/x}.         - Apply L'Hospital: lim⁡x→0+1/x−1/x2=lim⁡x→0+−x=0\lim_{x \rightarrow 0^+} \frac{1/x}{-1/x^2} = \lim_{x \rightarrow 0^+} -x = 0.         - If lim⁡ln⁡(y)=0\lim \ln(y) = 0, then lim⁡y=e0=1\lim y = e^0 = 1.