\[ \frac{d}{dx}f(x) =\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{(x+h)-x} = \lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h} \]
The Derivative as a Slope
\[\left[k f(x)\right]' = k f'(x)\]
\[\left[f(x)\pm g(x)\right]' = f'(x)\pm g'(x)\]
\[\left[f(x)g(x)\right]' = f'(x)g(x)+f(x)g'(x)\]
\(\frac{f(x)}{g(x)}' = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}, g(x)\neq 0\)$
\[\left[x^k\right]' = k x^{k-1}\]
\[f \circ g=f[g(x)]\]
\(g(x)=x^2\) for \(-\infty<x<\infty\)
\[f\circ g=\log x^2, -\infty<x<\infty\]
\[g\circ f = [\log x]^2, 0<x<\infty\]
\[y=f\circ g= f[g(x)]\]
\[\frac{d}{dx} \{ f[g(x)] \} = f'[g(x)] g'(x)\]
\[(e^x)^\prime = e^x\]
\[\log(x)^\prime = \frac{1}{x}\]
Composite functions
\[\left(e^{g(x)}\right)^\prime = e^{g(x)} \cdot g^\prime(x)\]
\[\left(\log g(x)\right)^\prime = \frac{g^\prime(x)}{g(x)}, ~~\text{if}~~ g(x) > 0\]
\[ \begin{eqnarray} \frac{d}{dx}f(x) & = & \lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ & = & \lim\limits_{h\to 0} \frac{e^{x + h} - e^x}{h} \\ & = & \lim\limits_{h\to 0} \frac{e^x e^h - e^x}{h} \\ & = & \lim\limits_{h\to 0} \frac{e^x(e^h - 1)}{h} \\ & = & e^x \lim\limits_{h\to 0} \frac{e^h - 1}{h} \end{eqnarray} \]
\[ \begin{eqnarray} \frac{d}{dx}f(x) & = & e^x \lim\limits_{h\to 0} \frac{e^h - 1}{h} \\ & = & e^x (1) \\ & = & e^x \end{eqnarray} \]
Derivative of the Exponential Function
Derivative of the Natural Log
Antiderivative of \(f(x) = \frac{1}{x^2}\)?
\[ \begin{eqnarray} \int \frac{1}{x^2} dx & = & -\frac{1}{x} + c \\ \frac{d}{dx} \left[ -\frac{1}{x} + c \right] & = & \frac{1}{x^2} \\ F(x) & = & \int f(x) dx \end{eqnarray} \]
\[f(x) = (x^2-4)\]
The Many Indefinite Integrals of a Function
Determine the area \(A(R)\) of a region \(R\) defined by a curve \(f(x)\) and some interval \(a\le x \le b\)
\[S(f,\Delta x)=\sum\limits_{i=1}^n f(x_i)\Delta x\]
Riemann sum
\[A(R)=\lim\limits_{\Delta x\to 0}\sum\limits_{i=1}^n f(x_i)\Delta x\]
\[\lim\limits_{\Delta x\to 0} \sum\limits_{i=1}^n f(x_i)\Delta x = \int\limits_a^b f(x) dx\]
\[\int\limits_a^b f(x) dx\]
\[F(x)=\int\limits_a^x f(t)dt, \quad a\le x\le b\]
\[F^\prime(x)=f(x), \quad a<x<b\]
\[\int\limits_a^bf(x)dx = F(b)-F(a)\]
Solve \(\int\limits_1^3 3x^2 dx\)
\[ \begin{eqnarray} f(x) & = & 3x^2 \\ F(x) & = & x^3 + c \\ \int\limits_1^3 3x^2 dx & = & F(3) - F(1) \\ & = & (3^3 + c) - (1^3 + c) \\ & = & 27 + c - 1 - c \\ & = & 26 \end{eqnarray} \]
Find a new function \(u(x)\) such that \(g(x)=f[u(x)]u'(x)\)
\[\int g(x) dx= \int f[u(x)]u'(x)dx = \int \frac{d}{dx} F[u(x)]dx = F[u(x)]+c\]
\[\int \frac{x}{x^2 + 1} dx\]
\[\frac{1}{2} \int \frac{2x}{x^2 + 1} dx\]
\[ \begin{eqnarray} u & = & x^2 + 1 \\ du & = & 2x \,dx \\ \int \frac{x}{x^2 + 1} dx & = & \frac{1}{2} \int \frac{2x}{x^2 + 1} dx \\ & = & \frac{1}{2} \int \frac{1}{u} du \\ & = & \frac{1}{2} \log(u) + c \\ & = & \frac{1}{2} \log(x^2 + 1) + c \\ \end{eqnarray} \]
Product rule
\[ \begin{eqnarray} \frac{d}{dx} f(x) g(x) & = & f(x) g'(x) + g(x) f'(x) \\ \frac{d}{dx}(uv) & = &u\frac{dv}{dx}+v\frac{du}{dx} \end{eqnarray} \]
Integration by parts
\[\int u dv = u v - \int v du\]
\[\int x \log(x) dx\]
\[ \begin{eqnarray} u &=& \log(x) \\ du &=& \frac{1}{x} \,dx \\ dv & = & x \, dx \\ v &=& \frac{1}{2}x^2 \end{eqnarray} \]
\[ \begin{eqnarray} \int x \log(x) \, dx &=& u v - \int v du \\ &=& \log(x) \times \frac{1}{2} x^2 - \int (\frac{1}{2} x^2) (\frac{1}{x} \, dx) \\ &=& \frac{1}{2} x^2 \log(x) - \int \frac{1}{2}x \,dx \\ &=& \frac{1}{2} x^2 \log(x) - \frac{1}{2} \int x \,dx \\ &=& \frac{1}{2} x^2 \log(x) - \frac{1}{2} \left( \frac{1}{2}x^2 \right) + c \\ &=& \frac{1}{2} x^2 \log(x) - \frac{1}{4} x^2 + c \\ \end{eqnarray} \]