Derivative

Proofs of the Derivative Rules

$\sin x$

$$\begin{array}{rl}& \underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}\\ =& \underset{h\to 0}{lim}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ =& \underset{h\to 0}{lim}\frac{\sin (x)\cos (h-1)+\cos x\sin h}{h}\\ =& \sin (x)\underset{h\to 0}{lim}\frac{\cos (h-1)}{h}+\cos (x)\underset{h\to 0}{lim}\frac{\sin h}{h}\end{array}$$$$\underset{h\to 0}{lim}\frac{\cos (h-1)}{h}=0$$$$\underset{h\to 0}{lim}\frac{\sin h}{h}=1$$$$\begin{array}{rl}& \sin (x)\underset{h\to 0}{lim}\frac{\cos (h-1)}{h}+\cos (x)\underset{h\to 0}{lim}\frac{\sin h}{h}\\ =& \cos x\end{array}$$

$\ln x$

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\ln x& =\underset{h\to 0}{lim}\frac{\ln (x+h)-\ln x}{h}\\ & =\underset{h\to 0}{lim}\frac{\ln (\frac{x+h}{x})}{h}\\ & =\underset{h\to 0}{lim}\frac{\ln (\frac{x}{x}+\frac{h}{x})}{h}\\ & =\underset{h\to 0}{lim}\frac{\ln (1+\frac{h}{x})}{h}\\ & =\underset{h\to 0}{lim}\ln (1+\frac{h}{x}{)}^{\frac{1}{h}}\end{array}$$

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

$$\begin{array}{rl}e& =\underset{h\to \mathrm{\infty }}{lim}(1+\frac{1}{h}{)}^h\\ e& =\underset{h\to 0}{lim}(1+h{)}^{\frac{1}{h}}\\ & \text{Change variable }h\text{ to }h/x\text{, }\\ & x\text{ is a nonzero constant.}\\ e& =\underset{h/x\to 0}{lim}(1+\frac{h}{x}{)}^{\frac{x}{h}}\\ e& =\underset{h\to 0}{lim}(1+\frac{h}{x}{)}^{\frac{x}{h}}\\ e^{\frac{1}{x}}& =\underset{h\to 0}{lim}(1+\frac{h}{x}{)}^{\frac{x}{h}\cdot \frac{1}{x}}\\ e^{\frac{1}{x}}& =\underset{h\to 0}{lim}(1+\frac{h}{x}{)}^{\frac{1}{h}}\end{array}$$$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\ln x=& \underset{h\to 0}{lim}\ln (1+\frac{h}{x}{)}^{\frac{1}{h}}\\ =& \underset{h\to 0}{lim}\ln e^{\frac{1}{x}}\\ =& \underset{h\to 0}{lim}\frac{1}{x}\\ =& \frac{1}{x}\end{array}$$

$(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$

$$\begin{array}{rl}(fg{)}^{\prime }& =\underset{h\to 0}{lim}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)g(x+h)-f(x)g(x)-f(x+h)g(x)+f(x+h)g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)(g(x+h)-g(x))+g(x)(f(x+h)-f(x))}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)(g(x+h)-g(x))}{h}+\underset{h\to 0}{lim}\frac{g(x)(f(x+h)-f(x))}{h}\\ & =\underset{h\to 0}{lim}(f(x+h)\frac{(g(x+h)-g(x))}{h})+\underset{h\to 0}{lim}(g(x)\frac{(f(x+h)-f(x))}{h})\\ & =(\underset{h\to 0}{lim}f(x+h))\left(\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}\right)+(\underset{h\to 0}{lim}g(x))\left(\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\right)\\ & =f(x)g^{\prime }(x)+f^{\prime }(x)g(x)\end{array}$$

$x^n$

$$\begin{array}{rl}y& =x^n\\ \ln y& =\ln x^n\\ \ln y& =n\ln x\\ \ln y& =n\ln x\\ \frac{\mathrm{d}}{\mathrm{d}x}(\ln y)& =\frac{\mathrm{d}}{\mathrm{d}x}(n\ln x)\\ \frac{y^{\prime }}{y}& =\frac{n}{x}\\ y^{\prime }& =y\frac{n}{x}\\ y^{\prime }& =x^n\frac{n}{x}\\ y^{\prime }& =n{x}^{n-1}\end{array}$$

Series

Misc

Differentiation of Geometric Series

$$\sum _{n=0}^{\mathrm{\infty }}{q}^n=\frac{1}{1-q},|q|<1$$

The first derivative:

$$\begin{array}{rl}\frac{d}{dq}\left(\sum _{n=0}^{\mathrm{\infty }}{q}^n\right)& =\sum _{n=1}^{\mathrm{\infty }}n{q}^{n-1}\\ & =\frac{d}{dq}\left(\frac{1}{1-q}\right)\\ & =\frac{0-1\times -1}{(1-q{)}^2}\\ & =\frac{1}{(1-q{)}^2}\end{array}$$$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}n{q}^{n-1}& =\frac{1}{(1-q{)}^2}\\ \sum _{n=1}^{\mathrm{\infty }}n{q}^{n-1}\cdot q& =\frac{q}{(1-q{)}^2}\\ \sum _{n=1}^{\mathrm{\infty }}n{q}^n& =\frac{q}{(1-q{)}^2}\end{array}$$

The second derivative:

$$\begin{array}{rl}\frac{d}{dq}\left(\sum _{n=0}^{\mathrm{\infty }}n{q}^n\right)& =\sum _{n=1}^{\mathrm{\infty }}{n}^2{q}^{n-1}\\ & =\frac{d}{dq}\left(\frac{q}{(1-q{)}^2}\right)\\ & =\frac{1\cdot (1-q{)}^2-q\cdot -1\cdot 2(1-q)}{(1-q{)}^4}\\ & =\frac{1-2q+q^2+2q-2q^2}{(1-q{)}^4}\\ & =\frac{1-q^2}{(1-q{)}^4}\end{array}$$$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}{n}^2{q}^{n-1}& =\frac{1-q^2}{(1-q{)}^4}\\ \sum _{n=1}^{\mathrm{\infty }}{n}^2{q}^n& =\frac{q(1+q)(1-q)}{(1-q{)}^4}\\ & =\frac{q(1+q)}{(1-q{)}^3}\end{array}$$