Chapter 6 | Inverse Functions

$$\begin{equation} \begin{split} f(x) &= a^x \\ f^{\prime}(x) &= a^x \cdot \ln(a) \\ \end{split} \end{equation}$$

$$\begin{equation} \begin{split} f(x) &= a^{g(x)} \\ f^{\prime}(x) &= a^{g(x)} \cdot \ln(a) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\Big(g(x)\Big) \\\\ &= a^{g(x)} \cdot \ln(a) \cdot g^{\prime}(x) \\ \end{split} \end{equation}$$

$$ \newcommand{\d}[1]{\frac{\mathrm{d}}{\mathrm{d}{#1}}} \newcommand{\df}[2]{\frac{\mathrm{d}{#1}}{\mathrm{d}{#2}}} \begin{equation} \begin{aligned} \overbrace{f(x) = y = a^x}^{\text{apply $\ln$ to both sides}} \\ \ln y = \ln b^x \\ \overbrace{\ln y = x \cdot \ln a \\}^{\text{differentiate}} \\ \overbrace{\frac{1}{y} \df{y}{x} = \cdot 1 \ln a}^{\text{multiply both sides with $y$}} \\ \overbrace{\df{y}{x} = \ln a \cdot y}^{\text{plugin original $y$}} \\ \overbrace{\df{y}{x} = \ln a \cdot a^x}^{\text{answer}} \\\\ f^{\prime}(x) = \ln a \cdot a^x \\ \end{aligned} \end{equation} $$

$$\begin{equation} \begin{split} f(x) &= \log_{b}(x) \\\\ f^{\prime}(x) &= \frac{1}{x \cdot \ln(b)} \end{split} \end{equation}$$

$$\begin{equation} \begin{split} f(x) &= \log_{b}\left(g(x)\right) \\\\ f^{\prime}(x) &= \frac{1}{g(x) \cdot \ln(b)} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\Big(g(x)\Big) \\\\ &= \frac{1}{g(x) \cdot \ln(b)} \cdot g^{\prime}(x) \end{split} \end{equation}$$

$$\begin{equation} \begin{split} f(x) &= a^x \\ f^{\prime}(x) &= a^x \cdot \ln(a) \\ \end{split} \end{equation}$$

$$\begin{equation} \begin{split} f(x) &= a^{g(x)} \\ f^{\prime}(x) &= a^{g(x)} \cdot \ln(a) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\Big(g(x)\Big) \\\\ &= a^{g(x)} \cdot \ln(a) \cdot g^{\prime}(x) \\ \end{split} \end{equation}$$

$$ \newcommand{\d}[1]{\frac{\mathrm{d}}{\mathrm{d}{#1}}} \newcommand{\df}[2]{\frac{\mathrm{d}{#1}}{\mathrm{d}{#2}}} \begin{equation} \begin{aligned} \overbrace{f(x) = y = a^x}^{\text{apply $\ln$ to both sides}} \\ \ln y = \ln b^x \\ \overbrace{\ln y = x \cdot \ln a \\}^{\text{differentiate}} \\ \overbrace{\frac{1}{y} \df{y}{x} = \cdot 1 \ln a}^{\text{multiply both sides with $y$}} \\ \overbrace{\df{y}{x} = \ln a \cdot y}^{\text{plugin original $y$}} \\ \overbrace{\df{y}{x} = \ln a \cdot a^x}^{\text{answer}} \\\\ f^{\prime}(x) = \ln a \cdot a^x \\ \end{aligned} \end{equation} $$

$$ \newcommand{\d}[1]{\frac{\mathrm{d}}{\mathrm{d}{#1}}} \newcommand{\df}[2]{\frac{\mathrm{d}{#1}}{\mathrm{d}{#2}}} \begin{equation} \begin{split} f(x) &= \log_{b}(x) \\\\ &= \overbrace{\frac{\ln(x)}{\ln(b)}}^{\text{change of base}} \\\\ &= \overbrace{\frac{1}{\ln(b)}}^{\text{constant}} \cdot \ln(x) \\\\ & \text{given}\\ & \d{x} \; \ln(x) = \frac{1}{x} \\\\ &\text{therefore} \\ f^{\prime}(x) &= \frac{1}{\ln(b)} \cdot \d{x}\Big(\ln(x) \Big) \\ &= \frac{1}{\ln(b)} \cdot \frac{1}{x} \\ &= \frac{1}{x \cdot \ln b} \end{split} \end{equation} $$

$$ e^{2\ln(3)} = 3^2 = 9 $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \underbrace{ \begin{aligned} \ln(1 + e^{-x}) &= 3 \\ e^3 &= 1 + e^{-x} \\ e^3 - 1 &= e^{-x} \end{aligned} }_{\text{Apply $\ln$ to both sides}} \end{equation} $$

$$ \begin{equation} \begin{split} \ln \left(e^3 - 1\right) &= \ln\left(e^{-x}\right) \\ \ln \left(e^3 - 1\right) &= -x \cdot \ln\left(e\right) \\ \ln \left(e^3 - 1\right) &= -x \cdot (1) \\ \ln \left(e^3 - 1\right) &= -x \\ -\ln \left(e^3 - 1\right) &= x \end{split} \end{equation} $$

$$ \begin{equation} x = -\ln \left(e^3 - 1\right) \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \underbrace{ \begin{aligned} \lnp{x + 1} + \lnp{x - 1} &= 1 \\ \lnp{(x + 1)(x - 1)} &= 1 \\ \lnp{x^2 - 1} &= 1 \\ e^1 &= x^2 - 1 \\ 1 + e^1 &= x^2 \\ \sqrt{1 + e^1} &= \sqrt{x^2} \\ \pm \sqrt{1 + e^1} &= x \\ \sqrt{1 + e^1} &= x \\ \end{aligned} }_{\text{no $\pm$ because of domain}} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} x = \sqrt{1 + e^1} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} tan^{-1}(x) &= 1 \\ x &= \frac{1}{8}\tau = \frac{1}{4}\pi \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} \sin(x) &= 0.3 = \frac{3}{10} \\ x = \arcsin(\frac{3}{10}) \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} f(x) = y &= \ln |\sec(5x) + \tan(5x)| \\\\ f^\prime(x) &= \frac{1}{\sec(5x) + \tan(5x)} \cdot \Big(\sec(5x) + \tan(5x)\Big)^{\prime} \\\\ &= \frac{\frac{5\sin(5x)}{\cos^2(x)} + 5\sec^2(5x)}{\sec(5x) + \tan(5x)} \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} \dox \sec(5x) &= \dox \frac{1}{\cos(5x)} \\ &= \dox \frac{A}{B} \\ &= \frac{A^{\prime} B - A B^{\prime}}{B^2} \\ &= \frac{0\cos(5x) - (-5\sin(5x))}{\cos^2(x)} \\ &= \frac{5\sin(5x)}{\cos^2(x)} \\ \dox \tan(5x) &= 5\sec^2(5x) \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} f^\prime(x) = \frac{\frac{5\sin(5x)}{\cos^2(x)} + 5\sec^2(5x)}{\sec(5x) + \tan(5x)} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} f(x) = y &= x \cdot \tan^{-1}(4x) \\ &= A \cdot B \\ f^{\prime}(x) &= A^{\prime} B + A B^{\prime} \\ &= \tan^{-1}(4x) + x \frac{1}{1 + (4x)^2} \cdot \dox\Big(4x\Big) \\ &= \tan^{-1}(4x) + \frac{4x}{1 + 16x^2} \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} f^\prime(x) = \tan^{-1}(4x) + \frac{4x}{1 + 16x^2} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} f(x) = y &= e^{mx} \cdot \cos(n x) \\ &= AB \\ f^{\prime}(x) &= \dox AB \\ &= A^{\prime} B + A B^{\prime} \\ &= \left(m e^{mx}\right) \cos(n x) + e^{mx} -\sin(n x) \cdot \dox\Big(n x\Big) \\ &= m e^{mx}\cos(n x) - n e^{mx}\sin(n x) \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} f^\prime(x) = m e^{mx}\cos(n x) - n e^{mx}\sin(n x) \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} f(x) = y &= 3^{x \cdot \ln(x)} \\ f^\prime(x) &= 3^{x \cdot \ln(x)} \cdot \ln(3) \cdot \dox\Big(x \cdot \ln(x)\Big) \\ &= 3^{x \cdot \ln(x)} \cdot \ln(3) \cdot \Big(\ln(x) + 1\Big) \\ &= \ln(3) \cdot 3^{x \cdot \ln(x)} \cdot \Big(\ln(x) + 1\Big) \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} \dox\Big(x \cdot \ln(x)\Big) &= \cdot AB \\ &= A^{\prime} B + A B^{\prime} \\ &= \ln(x) + x \frac{1}{x} \\ &= \ln(x) + 1 \\ \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} f^\prime(x) = \ln(3) \cdot 3^{x \cdot \ln(x)} \cdot \Big(\ln(x) + 1\Big) \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to \infty} e^{-3x} = \lim_{x \to \infty} e^{u} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} u = -3x \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} x \to \infty, \;\; u \to -\infty \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to \infty} e^{-3x} = \lim_{u \to -\infty} e^u = 0 \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to -\infty} e^x = 0 \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to 10^{-}} \ln(100 - x^2) \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} u = 100 - x^2 \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} x \to 10, \;\; u \to 0 \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to 10^{-}} \ln(100 - x^2) = \lim_{u \to 0} \ln(u) = -\infty \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to 0} \ln(x) = -\infty \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to \infty} \arctan(x^3 - x) \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \underbrace{u = x^3 - x = x(x^2 - 1) = x(x + 1)(x - 1)}_{\text{Odd - Increasing Function}} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} x \to \infty, \;\; u \to \infty \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to \infty} \arctan(x^3 - x) = \lim_{u \to \infty} \arctan(u) = \frac{\pi}{2} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to \infty} \arctan(x) = \frac{\pi}{2} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \newcommand{\eqWithMsg}[1]{\overset{{#1}}{=}} \begin{equation} \begin{split} \lim_{x \to 0} \frac{e^{4x} - 1 - 4x}{x^2} &\eqWithMsg{\frac{0}{0}} \lim_{x \to 0} \frac{4e^{4x} - 4}{2x} \\\\ &\eqWithMsg{\frac{0}{0}} \lim_{x \to 0} \frac{16e^{4x}}{2} \\\\ &= \frac{16}{2} \lim_{x \to 0} e^{4x} \\\\ &= 8 \lim_{x \to 0} e^{4x} \\\\ \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} u = 4x \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} x \to 0, \;\; u \to 0 \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \newcommand{\eqWithMsg}[1]{\overset{{#1}}{=}} \begin{equation} \begin{split} \lim_{x \to 0} e^{4x} &= \lim_{u \to 0} e^u \eqWithMsg{DSP} e^0 = 1 \\\\ 8\lim_{x \to 0} e^{4x} &= 8\lim_{u \to 0} e^u = 8(1) = 8 \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} \lim_{x \to \infty} \frac{e^{4x} - 1 - 4x}{x^2} &\eqWithMsg{\frac{0}{0}} \lim_{x \to \infty} \frac{4e^{4x} - 4}{2x} \\\\ &\eqWithMsg{\frac{0}{0}} \lim_{x \to \infty} \frac{16e^{4x}}{2} \\\\ &= \frac{16}{2} \lim_{x \to \infty} e^{4x} \\\\ &= 8 \lim_{x \to \infty} e^{4x} \\\\ \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} u = 4x \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} x \to \infty, \;\; u \to \infty \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} \lim_{x \to 0} e^{4x} &= \lim_{u \to 0} e^u = \infty \\\\ 8\lim_{x \to 0} e^{4x} &= 8\lim_{u \to 0} e^u = 8 \cdot \infty = \infty \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \lim_{x \to \infty} e^x = \infty \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} \lim_{x \to -\infty} \Big( x^2 - x^3 \Big) e^{2x} \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} u &= \overbrace{x^2 - x^3 = x^2(1 - x)}^{\text{Odd, Decreasing Function}} \\ v &= 2x \\ \end{split} \end{equation} $$

$$ \newcommand{\dox}{\frac{\mathrm{d}}{\mathrm{d}x}} \newcommand{\lnp}[1]{\ln\left({#1}\right)} \begin{equation} \begin{split} x \to -\infty, \;\; u \to \infty \\ x \to -\infty, \;\; v \to -\infty \\ \end{split} \end{equation} $$