Derivative

Proofs of the Derivative Rules

\begin{aligned} lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h} \\ = & lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\ = & lim_{h \to 0} \frac{\sin (x) \cos (h - 1) + \cos x \sin h}{h} \\ = & \sin (x) lim_{h \to 0} \frac{\cos (h - 1)}{h} + \cos (x) lim_{h \to 0} \frac{\sin h}{h} \end{aligned}

lim_{h \to 0} \frac{\cos (h - 1)}{h} = 0

lim_{h \to 0} \frac{\sin h}{h} = 1

\begin{aligned} \sin (x) lim_{h \to 0} \frac{\cos (h - 1)}{h} + \cos (x) lim_{h \to 0} \frac{\sin h}{h} \\ = & \cos x \end{aligned}

\begin{aligned} \frac{d}{d x} \ln x & = lim_{h \to 0} \frac{\ln (x + h) - \ln x}{h} \\ = lim_{h \to 0} \frac{\ln (\frac{x + h}{x})}{h} \\ = lim_{h \to 0} \frac{\ln (\frac{x}{x} + \frac{h}{x})}{h} \\ = lim_{h \to 0} \frac{\ln (1 + \frac{h}{x})}{h} \\ = lim_{h \to 0} \ln (1 + \frac{h}{x})^{\frac{1}{h}} \end{aligned}

Now, deal with $(1 + \frac{h}{x})^{\frac{1}{h}}$ :

\begin{aligned} e & = lim_{h \to \infty} (1 + \frac{1}{h})^{h} \\ e & = lim_{h \to 0} (1 + h)^{\frac{1}{h}} \\ Change variable h to h / x, \\ x is a nonzero constant. \\ e & = lim_{h / x \to 0} (1 + \frac{h}{x})^{\frac{x}{h}} \\ e & = lim_{h \to 0} (1 + \frac{h}{x})^{\frac{x}{h}} \\ e^{\frac{1}{x}} & = lim_{h \to 0} (1 + \frac{h}{x})^{\frac{x}{h} \cdot \frac{1}{x}} \\ e^{\frac{1}{x}} & = lim_{h \to 0} (1 + \frac{h}{x})^{\frac{1}{h}} \end{aligned}

\begin{aligned} \frac{d}{d x} \ln x = & lim_{h \to 0} \ln (1 + \frac{h}{x})^{\frac{1}{h}} \\ = & lim_{h \to 0} \ln e^{\frac{1}{x}} \\ = & lim_{h \to 0} \frac{1}{x} \\ = & \frac{1}{x} \end{aligned}

\begin{aligned} (f g)^{'} & = lim_{h \to 0} \frac{f (x + h) g (x + h) - f (x) g (x)}{h} \\ = lim_{h \to 0} \frac{f (x + h) g (x + h) - f (x) g (x) - f (x + h) g (x) + f (x + h) g (x)}{h} \\ = lim_{h \to 0} \frac{f (x + h) (g (x + h) - g (x)) + g (x) (f (x + h) - f (x))}{h} \\ = lim_{h \to 0} \frac{f (x + h) (g (x + h) - g (x))}{h} + lim_{h \to 0} \frac{g (x) (f (x + h) - f (x))}{h} \\ = lim_{h \to 0} (f (x + h) \frac{(g (x + h) - g (x))}{h}) + lim_{h \to 0} (g (x) \frac{(f (x + h) - f (x))}{h}) \\ = (lim_{h \to 0} f (x + h)) (lim_{h \to 0} \frac{g (x + h) - g (x)}{h}) + (lim_{h \to 0} g (x)) (lim_{h \to 0} \frac{f (x + h) - f (x)}{h}) \\ = f (x) g^{'} (x) + f^{'} (x) g (x) \end{aligned}

\begin{aligned} y & = x^{n} \\ \ln y & = \ln x^{n} \\ \ln y & = n \ln x \\ \ln y & = n \ln x \\ \frac{d}{d x} (\ln y) & = \frac{d}{d x} (n \ln x) \\ \frac{y^{'}}{y} & = \frac{n}{x} \\ y^{'} & = y \frac{n}{x} \\ y^{'} & = x^{n} \frac{n}{x} \\ y^{'} & = n x^{n - 1} \end{aligned}

\sum_{n = 0}^{\infty} q^{n} = \frac{1}{1 - q}, | q | < 1

The first derivative:

\begin{aligned} \frac{d}{d q} (\sum_{n = 0}^{\infty} q^{n}) & = \sum_{n = 1}^{\infty} n q^{n - 1} \\ = \frac{d}{d q} (\frac{1}{1 - q}) \\ = \frac{0 - 1 \times - 1}{(1 - q)^{2}} \\ = \frac{1}{(1 - q)^{2}} \end{aligned}

\begin{aligned} \sum_{n = 1}^{\infty} n q^{n - 1} & = \frac{1}{(1 - q)^{2}} \\ \sum_{n = 1}^{\infty} n q^{n - 1} \cdot q & = \frac{q}{(1 - q)^{2}} \\ \sum_{n = 1}^{\infty} n q^{n} & = \frac{q}{(1 - q)^{2}} \end{aligned}

The second derivative:

\begin{aligned} \frac{d}{d q} (\sum_{n = 0}^{\infty} n q^{n}) & = \sum_{n = 1}^{\infty} n^{2} q^{n - 1} \\ = \frac{d}{d q} (\frac{q}{(1 - q)^{2}}) \\ = \frac{1 \cdot (1 - q)^{2} - q \cdot - 1 \cdot 2 (1 - q)}{(1 - q)^{4}} \\ = \frac{1 - 2 q + q^{2} + 2 q - 2 q^{2}}{(1 - q)^{4}} \\ = \frac{1 - q^{2}}{(1 - q)^{4}} \end{aligned}

\begin{aligned} \sum_{n = 1}^{\infty} n^{2} q^{n - 1} & = \frac{1 - q^{2}}{(1 - q)^{4}} \\ \sum_{n = 1}^{\infty} n^{2} q^{n} & = \frac{q (1 + q) (1 - q)}{(1 - q)^{4}} \\ = \frac{q (1 + q)}{(1 - q)^{3}} \end{aligned}