$$ \begin{align} f(x) &= n\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{n-n}{h}\\ &= \lim_{h\to0}\frac{0}{h}\\ &= \lim_{h\to0}0\\ &= 0\\ \end{align} $$

$$ \begin{align} f(x) &= x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{(x+h)-x}{h}\\ &= \lim_{h\to0}\frac{h}{h}\\ &= \lim_{h\to0}1\\ &= 1\\ \end{align} $$

$$ \begin{align} f(x) &= x^2\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{(x+h)^2-x^2}{h}\\ &= \lim_{h\to0}\frac{(x^2+2hx+h^2)-x^2}{h}\\ &= \lim_{h\to0}\frac{2hx+h^2}{h}\\ &= \lim_{h\to0}2x+h\\ &= 2x\\ \end{align} $$

$$ \begin{align} \text{Define } C_k^n &= \prod_{r=1}^k\frac{n-r+1}{r}\text{, }\\ \text{then }f(x) &= x^n\text{.}\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{(x+h)^n-x^n}{h}\\ &= \lim_{h\to0}\frac{(\sum_{k=0}^{\infty}C_k^n x^{n-k} h^k)-x^n}{h}\\ &= \lim_{h\to0}\frac{(\sum_{k=1}^{\infty}C_k^n x^{n-k} h^k)+x^n-x^n}{h}\\ &= \lim_{h\to0}\sum_{k=1}^{\infty}C_k^n x^{n-k} h^{k-1}\\ &= \lim_{h\to0}(n x^{n-1}+\sum_{k=2}^{\infty}C_k^n x^{n-k} h^{k-1})\\ &= n x^{n-1}+\lim_{h\to0}\sum_{k=1}^{\infty}C_{k+1}^n x^{n-(k+1)} h^{k}\\ &= n x^{n-1}+0\\ &= n x^{n-1}\\ \end{align} $$

$$ \begin{align} f(x) &= \frac{1}{x}\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ &= \lim_{h\to0}\frac{x-(x+h)}{hx(x+h)}\\ &= \lim_{h\to0}\frac{-h}{hx(x+h)}\\ &= -\lim_{h\to0}\frac{1}{x(x+h)}\\ &= -\frac{1}{x\cdot x}\\ &= -\frac{1}{x^2}\\ \end{align} $$

$$ \begin{align} f(x) &= \frac{1}{x^2}\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\\ &= \lim_{h\to0}\frac{x^2-(x+h)^2}{hx^2(x+h)^2}\\ &= \lim_{h\to0}\frac{x^2-(x^2+2hx+h^2)}{hx^2(x+h)^2}\\ &= -\lim_{h\to0}\frac{2hx+h^2}{hx^2(x+h)^2}\\ &= -\lim_{h\to0}\frac{2x+h}{x^2(x+h)^2}\\ &= -\frac{2x}{x^2\cdot x^2}\\ &= -\frac{2}{x^3}\\ \end{align} $$

$$ \begin{align} \text{Define } C_k^n &= \prod_{r=1}^k\frac{n-r+1}{r}\text{, }\\ \text{then }f(x) &= \frac{1}{x^n}\text{.}\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\frac{1}{(x+h)^n}-\frac{1}{x^n}}{h}\\ &= -\lim_{h\to0}\frac{(x+h)^n-x^n}{hx^n(x+h)^n}\\ &= -\lim_{h\to0}\frac{(\sum_{k=0}^{\infty}C_k^n x^{n-k} h^k)-x^n}{hx^n(x+h)^n}\\ &= -\lim_{h\to0}\frac{(\sum_{k=1}^{\infty}C_k^n x^{n-k} h^k)+x^n-x^n}{hx^n(x+h)^n}\\ &= -\lim_{h\to0}\frac{\sum_{k=1}^{\infty}C_k^n x^{n-k} h^{k-1}}{x^n(x+h)^n}\\ &= -\lim_{h\to0}\frac{n x^{n-1}+\sum_{k=2}^{\infty}C_k^n x^{n-k} h^{k-1}}{x^n(x+h)^n}\\ &= -\lim_{h\to0}\frac{n x^{n-1}+\sum_{k=1}^{\infty}C_{k+1}^n x^{n-(k+1)} h^k}{x^n(x+h)^n}\\ &= -\frac{n x^{n-1}+\lim_{h\to0}\sum_{k=1}^{\infty}C_{k+1}^n x^{n-(k+1)} h^k}{x^n\cdot x^n}\\ &= -\frac{n x^{n-1}+0}{x^{2n}}\\ &= -\frac{n}{x^{n+1}}\\ \end{align} $$

$$ \begin{align} f(x) &= \sqrt{x}\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\\ &= \lim_{h\to0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\\ &= \lim_{h\to0}\frac{\sqrt{x+h}^2-\sqrt{x}^2}{h(\sqrt{x+h}+\sqrt{x})}\\ &= \lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\\ &= \lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ &= \lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &= \frac{1}{\sqrt{x}+\sqrt{x}}\\ &= \frac{1}{2\sqrt{x}}\\ \end{align} $$

$$ \begin{align} f(x) &= \sqrt[3]{x}\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}\\ &= \lim_{h\to0}\frac{(\sqrt[3]{x+h}-\sqrt[3]{x})(\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2)}{h(\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2)}\\ &= \lim_{h\to0}\frac{\sqrt[3]{x+h}^3-\sqrt[3]{x}^3}{h(\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2)}\\ &= \lim_{h\to0}\frac{(x+h)-x}{h(\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2)}\\ &= \lim_{h\to0}\frac{h}{h(\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2)}\\ &= \lim_{h\to0}\frac{1}{\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2}\\ &= \frac{1}{\sqrt[3]{x}^2+\sqrt[3]{x}^2+\sqrt[3]{x}^2}\\ &= \frac{1}{3\sqrt[3]{x}^2}\\ \end{align} $$

$$ \begin{align} f(x) &= \sin x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}\\ &= \lim_{h\to0}\frac{2\cos (x+\frac{h}{2})\sin \frac{h}{2}}{h}\\ &= \lim_{\frac{h}{2}\to0}\frac{\cos (x+\frac{h}{2})\sin \frac{h}{2}}{\frac{h}{2}}\\ &= \lim_{\frac{h}{2}\to0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\cdot\lim_{\frac{h}{2}\to0}\cos (x+\frac{h}{2})\\ &= 1\cdot\cos x\\ &= \cos x\\ \end{align} $$

$$ \begin{align} f(x) &= \cos x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}\\ &= -\lim_{h\to0}\frac{2\sin (x+\frac{h}{2})\sin \frac{h}{2}}{h}\\ &= -\lim_{\frac{h}{2}\to0}\frac{\sin (x+\frac{h}{2})\sin \frac{h}{2}}{\frac{h}{2}}\\ &= -\lim_{\frac{h}{2}\to0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\cdot\lim_{\frac{h}{2}\to0}\sin (x+\frac{h}{2})\\ &= -1\cdot\sin x\\ &= -\sin x\\ \end{align} $$

$$ \begin{align} f(x) &= \tan x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\tan(x+h)-\tan x}{h}\\ &= \lim_{h\to0}\frac{\frac{\sin(x+h)}{\cos(x+h)}-\frac{\sin x}{\cos x}}{h}\\ &= \lim_{h\to0}\frac{\sin(x+h)\cos x-\sin x\cos(x+h)}{h\cos x\cos(x+h)}\\ &= \lim_{h\to0}\frac{\sin(2x+h)+\sin h-\sin(2x+h)-\sin(-h)}{2h\cos x\cos(x+h)}\\ &= \lim_{h\to0}\frac{\sin h+\sin h}{2h\cos x\cos(x+h)}\\ &= \lim_{h\to0}\frac{\sin h}{h\cos x\cos(x+h)}\\ &= \lim_{h\to0}\frac{\sin h}{h}\cdot\lim_{h\to0}\frac{1}{\cos x\cos(x+h)}\\ &= 1\cdot\frac{1}{\cos x\cdot\cos x}\\ &= \frac{1}{\cos^2 x}\\ &= \sec^2 x\\ \end{align} $$

$$ \begin{align} f(x) &= \cot x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\cot(x+h)-\cot x}{h}\\ &= \lim_{h\to0}\frac{\frac{\cos(x+h)}{\sin(x+h)}-\frac{\cos x}{\sin x}}{h}\\ &= \lim_{h\to0}\frac{\cos(x+h)\sin x-\cos x\sin(x+h)} {h\sin x\sin(x+h)}\\ &= \lim_{h\to0}\frac{\sin(2x+h)+\sin (-h)-\sin(2x+h)-\sin h}{2h\sin x\sin(x+h)}\\ &= \lim_{h\to0}\frac{-\sin h-\sin h}{2h\sin x\sin(x+h)}\\ &= -\lim_{h\to0}\frac{\sin h}{h\sin x\sin(x+h)}\\ &= -\lim_{h\to0}\frac{\sin h}{h}\cdot\lim_{h\to0}\frac{1}{\sin x\sin(x+h)}\\ &= -1\cdot\frac{1}{\sin x\cdot\sin x}\\ &= -\frac{1}{\sin^2 x}\\ &= -\csc^2 x\\ \end{align} $$

$$ \begin{align} f(x) &= \sec x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\sec(x+h)-\sec x}{h}\\ &= \lim_{h\to0}\frac{\frac{1}{\cos(x+h)}-\frac{1}{\cos x}}{h}\\ &= \lim_{h\to0}\frac{\cos x-\cos (x+h)} {h\cos x\cos(x+h)}\\ &= \lim_{h\to0}\frac{2\sin (x+\frac{h}{2})\sin \frac{h}{2}} {h\cos x\cos(x+h)}\\ &= \frac{1}{\cos x}\cdot\lim_{\frac{h}{2}\to0}\frac{\sin\frac{h}{2} }{\frac{h}{2}}\cdot\lim_{h\to0}\frac{\sin (x+\frac{h}{2})} {\cos(x+h)}\\ &= \frac{1}{\cos x}\cdot 1\cdot\frac{\sin x}{\cos x}\\ &= \sec x\tan x\\ \end{align} $$

$$ \begin{align} f(x) &= \csc x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\csc(x+h)-\csc x}{h}\\ &= \lim_{h\to0}\frac{\frac{1}{\sin(x+h)}-\frac{1}{\sin x}}{h}\\ &= \lim_{h\to0}\frac{\sin x-\sin (x+h)} {h\sin x\sin(x+h)}\\ &= \lim_{h\to0}\frac{-2\cos (x+\frac{h}{2})\sin \frac{h}{2}} {h\sin x\sin(x+h)}\\ &= -\frac{1}{\sin x}\cdot\lim_{\frac{h}{2}\to0}\frac{\sin\frac{h}{2} }{\frac{h}{2}}\cdot\lim_{h\to0}\frac{\cos (x+\frac{h}{2})} {\sin(x+h)}\\ &= -\frac{1}{\sin x}\cdot 1\cdot\frac{\cos x}{\sin x}\\ &= -\csc x\cot x\\ \end{align} $$

$$ \begin{align} f(x) &= e^x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{e^{x+h}-e^x}{h}\\ &= e^x\lim_{h\to0}\frac{e^h-1}{h}\\ &= e^x\cdot 1\\ &= e^x\\ \end{align} $$

$$ \begin{align} f(x) &= n^x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{n^{x+h}-n^x}{h}\\ &= n^x\lim_{h\to0}\frac{n^h-1}{h}\\ &= n^x\lim_{h\to0}\frac{(e^{\ln n})^h-1}{h}\\ &= n^x\lim_{h\to0}\frac{(e^{h \ln n})-1}{h}\\ &= n^x\ln n\lim_{h\ln n\to0}\frac{(e^{h \ln n})-1}{h \ln n}\\ &= n^x\ln n\cdot 1\\ &= n^x\ln n\\ \end{align} $$

$$ \begin{align} f(x) &= \ln x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\ln (x+h)-\ln x}{h}\\ &= \lim_{h\to0}\frac{\ln \frac{x+h}{x}}{h}\\ &= \frac{1}{x}\lim_{h\to0}\frac{\ln (1+\frac{h}{x})}{\frac{h}{x}}\\ &= \frac{1}{x}\lim_{h\to0}\frac{x}{h}\ln (1+\frac{h}{x})\\ &= \frac{1}{x}\lim_{h\to0}\ln (1+\frac{h}{x})^{\frac{x}{h}}\\ &= \frac{1}{x}\ln\lim_{\frac{x}{h}\to\infty} (1+\frac{1}{\frac{x}{h}})^{\frac{x}{h}}\\ &= \frac{1}{x}\ln e\\ &= \frac{1}{x}\cdot 1\\ &= \frac{1}{x}\\ \end{align} $$

$$ \begin{align} f(x) &= \log_n x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\log_n (x+h)-\log_n x}{h}\\ &= \lim_{h\to0}\frac{\log_n \frac{x+h}{x}}{h}\\ &= \lim_{h\to0}\frac{\ln \frac{x+h}{x}}{h\ln n}\\ &= \frac{1}{x\ln n}\lim_{h\to0}\frac{\ln (1+\frac{h}{x})}{\frac{h}{x}}\\ &= \frac{\log_n e}{x\log_n n}\lim_{h\to0}\frac{x}{h}\ln (1+\frac{h}{x})\\ &= \frac{\log_n e}{x\cdot 1}\lim_{h\to0}\ln (1+\frac{h}{x})^{\frac{x}{h}}\\ &= \frac{\log_n e}{x}\ln\lim_{\frac{x}{h}\to\infty} (1+\frac{1}{\frac{x}{h}})^{\frac{x}{h}}\\ &= \frac{\log_n e}{x}\ln e\\ &= \frac{\log_n e}{x}\cdot 1\\ &= \frac{\log_n e}{x}\\ \end{align} $$

$$ \begin{align} \text{Define } C_k^n &= \prod_{r=1}^k\frac{n-r+1}{r}\text{, }\\ \text{then }f(x) &= x^x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{(x+h)^{x+h}-x^x}{h}\\ &= \color{red}\lim_{h\to0}\frac{(x+h)^{x+h}-(x+h)^x}{h}\color{black}+\color{blue}\lim_{h\to0}\frac{(x+h)^x-x^x}{h}\\ &= \color{red}\lim_{h\to0}(x+h)^x\cdot\lim_{h\to0}\frac{(x+h)^h-1}{h}\color{black}+\color{blue}\lim_{h\to0}\frac{(\sum_{k=0}^{\infty}C_k^x x^{x-k} h^k)-x^x}{h}\\ &= \color{red}\lim_{h\to0}(x+h)^x\cdot\lim_{h\to0}\frac{(e^{\ln(x+h)})^h-1}{h}\color{black}+\color{blue}\lim_{h\to0}\frac{(\sum_{k=1}^{\infty}C_k^x x^{x-k} h^k)+x^x-x^x}{h}\\ &= \color{red}x^x\lim_{h\to0}\ln(x+h)\cdot\lim_{h\ln(x+h)\to0}\frac{(e^{h\ln(x+h)})-1}{h\ln(x+h)}\color{black}+\color{blue}\lim_{h\to0}\sum_{k=1}^{\infty}C_k^x x^{x-k} h^{k-1}\\ &= \color{red}x^x\ln x\cdot1\color{black}+\color{blue}\lim_{h\to0}(x \cdot x^{x-1}+\sum_{k=2}^{\infty}C_k^x x^{x-k} h^{k-1})\\ &= \color{red}x^x\ln x\color{black}+\color{blue}x^x+\lim_{h\to0}\sum_{k=1}^{\infty}C_{k+1}^x x^{x-(k+1)} h^{k}\\ &= x^x(\ln x+1)+0\\ &= x^x(\ln x+1)\\ \end{align} $$

$$ \begin{align} f(x) &= \sin^{-1}x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{h}\\ &= \lim_{h\to0}\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{x+h-x}\\ &= \lim_{h\to0}\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{\sin(\sin^{-1}(x+h))-\sin(\sin^{-1}x)}\\ &= \lim_{h\to0}\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2\cos\frac{\color{blue}\sin^{-1}(x+h)+\sin^{-1}x}{2}\sin\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2}}\\ &= \lim_{h\to0}\frac{\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2}}{\cos\frac{\color{blue}\sin^{-1}(x+h)+\sin^{-1}x}{2}\sin\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2}}\\ &= \frac{1}{\lim_{h\to0}\cos\frac{\color{blue}\sin^{-1}(x+h)+\sin^{-1}x}{2}\lim_{\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2}\to0}\frac{\sin\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2}}{\frac{\color{red}\sin^{-1}(x+h)-\sin^{-1}x}{2}}}\\ &= \frac{1}{\cos\frac{\sin^{-1}x+\sin^{-1}x}{2}\cdot 1}\\ &= \frac{1}{\cos(\sin^{-1}x)}&(\frac{\pi}{2}\geq\sin^{-1}x\geq-\frac{\pi}{2})\\ &= \frac{1}{\sqrt{1-(\sin(\sin^{-1}x))^2}}&(\cos(\sin^{-1}x)\geq0)\\ &= \frac{1}{\sqrt{1-x^2}}\\ \end{align} $$

$$ \begin{align} f(x) &= \cos^{-1}x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{h}\\ &= \lim_{h\to0}\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{x+h-x}\\ &= \lim_{h\to0}\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{\cos(\cos^{-1}(x+h))-\cos(\cos^{-1}x)}\\ &= \lim_{h\to0}\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{-2\sin\frac{\color{blue}\cos^{-1}(x+h)+\cos^{-1}x}{2}\sin\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{2}}\\ &= -\lim_{h\to0}\frac{\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{2}}{\sin\frac{\color{blue}\cos^{-1}(x+h)+\cos^{-1}x}{2}\sin\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{2}}\\ &= -\frac{1}{\lim_{h\to0}\sin\frac{\color{blue}\cos^{-1}(x+h)+\cos^{-1}x}{2}\lim_{\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{2}\to0}\frac{\sin\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{2}}{\frac{\color{red}\cos^{-1}(x+h)-\cos^{-1}x}{2}}}\\ &= -\frac{1}{\sin\frac{\cos^{-1}(x)+\cos^{-1}x}{2}\cdot 1}\\ &= -\frac{1}{\sin(\cos^{-1}x)}&(\pi\geq\cos^{-1}x\geq0)\\ &= -\frac{1}{\sqrt{1-(\cos(\cos^{-1}x))^2}}&(\sin(\cos^{-1}x)\geq0)\\ &= -\frac{1}{\sqrt{1-x^2}}\\ \end{align} $$

$$ \begin{align} f(x) &= \tan^{-1}x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\color{red}\tan^{-1}(x+h)-\tan^{-1}x}{h}\\ &= \lim_{h\to0}\frac{\color{red}\tan^{-1}(x+h)-\tan^{-1}x}{x+h-x}\\ &= \lim_{h\to0}\frac{\color{red}\tan^{-1}(x+h)-\tan^{-1}x}{\tan(\tan^{-1}(x+h))-\tan(\tan^{-1}x)}\\ &= \lim_{h\to0}\frac{\color{red}\tan^{-1}(x+h)-\tan^{-1}x}{\frac{\sin(\tan^{-1}(x+h))}{\cos(\tan^{-1}(x+h))}-\frac{\sin(\tan^{-1}x)}{\cos(\tan^{-1}x)}}\\ &= \lim_{h\to0}\frac{\color{red}\tan^{-1}(x+h)-\tan^{-1}x}{\frac{\sin(\tan^{-1}(x+h))\cos(\tan^{-1}x)-\sin(\tan^{-1}x)\cos(\tan^{-1}(x+h))}{\cos(\tan^{-1}(x+h))\cos(\tan^{-1}x)}}\\ &= \lim_{h\to0}\frac{(\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})(\cos(\tan^{-1}(x+h))\cos(\tan^{-1}x))}{\sin(\tan^{-1}(x+h))\cos(\tan^{-1}x)-\sin(\tan^{-1}x)\cos(\tan^{-1}(x+h))}\\ &= \lim_{h\to0}\frac{(\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})(\cos(\tan^{-1}(x+h))\cos(\tan^{-1}x))}{\frac{1}{2}(\sin (\color{blue}\tan^{-1}(x+h)+\tan^{-1}x\color{black})+\sin (\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})-\sin (\color{blue}\tan^{-1}(x+h)+\tan^{-1}x\color{black})-\sin (-(\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})))}\\ &= \lim_{h\to0}\frac{(\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})(\cos(\tan^{-1}(x+h))\cos(\tan^{-1}x))}{\frac{1}{2}(2\sin (\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black}))}\\ &= \lim_{h\to0}\frac{(\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})(\cos(\tan^{-1}(x+h))\cos(\tan^{-1}x))}{\sin (\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})}\\ &= \lim_{\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black}\to0}\frac{\color{red}\tan^{-1}(x+h)-\tan^{-1}x}{\sin (\color{red}\tan^{-1}(x+h)-\tan^{-1}x\color{black})}\cdot\lim_{h\to0}(\cos(\tan^{-1}(x+h))\cos(\tan^{-1}x))\\ &= 1\cdot(\cos(\tan^{-1}x)\cdot\cos(\tan^{-1}x))\\ &= \cos^2(\tan^{-1}x)\\ &= \frac{1}{\sec^2(\tan^{-1}x)}\\ &= \frac{1}{1+(\tan(\tan^{-1}x))^2}\\ &= \frac{1}{1+x^2}\\ \end{align} $$

$$ \begin{align} f(x) &= \cot^{-1}x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\color{red}\cot^{-1}(x+h)-\cot^{-1}x}{h}\\ &= \lim_{h\to0}\frac{\color{red}\cot^{-1}(x+h)-\cot^{-1}x}{x+h-x}\\ &= \lim_{h\to0}\frac{\color{red}\cot^{-1}(x+h)-\cot^{-1}x}{\cot(\cot^{-1}(x+h))-\cot(\cot^{-1}x)}\\ &= \lim_{h\to0}\frac{\color{red}\cot^{-1}(x+h)-\cot^{-1}x}{\frac{\cos(\cot^{-1}(x+h))}{\sin(\cot^{-1}(x+h))}-\frac{\cos(\cot^{-1}x)}{\sin(\cot^{-1}x)}}\\ &= \lim_{h\to0}\frac{\color{red}\cot^{-1}(x+h)-\cot^{-1}x}{\frac{\cos(\cot^{-1}(x+h))\sin(\cot^{-1}x)-\cos(\cot^{-1}x)\sin(\cot^{-1}(x+h))}{\sin(\cot^{-1}(x+h))\sin(\cot^{-1}x)}}\\ &= \lim_{h\to0}\frac{(\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black})(\sin(\cot^{-1}(x+h))\sin(\cot^{-1}x))}{\cos(\cot^{-1}(x+h))\sin(\cot^{-1}x)-\cos(\cot^{-1}x)\sin(\cot^{-1}(x+h))}\\ &= \lim_{h\to0}\frac{(\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black})(\sin(\cot^{-1}(x+h))\sin(\cot^{-1}x))}{\frac{1}{2}(\sin (\color{blue}\cot^{-1}(x+h)+\cot^{-1}x\color{black})+\sin (-(\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black}))-\sin (\color{blue}\cot^{-1}(x+h)+\cot^{-1}x\color{black})-\sin (\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black}))}\\ &= \lim_{h\to0}\frac{(\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black})(\sin(\cot^{-1}(x+h))\sin(\cot^{-1}x))}{\frac{1}{2}(-2\sin (\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black}))}\\ &= -\lim_{h\to0}\frac{(\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black})(\sin(\cot^{-1}(x+h))\sin(\cot^{-1}x))}{\sin (\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black})}\\ &= -\lim_{\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black}\to0}\frac{\color{red}\cot^{-1}(x+h)-\cot^{-1}x}{\sin (\color{red}\cot^{-1}(x+h)-\cot^{-1}x\color{black})}\cdot\lim_{h\to0}(\sin(\cot^{-1}(x+h))\sin(\cot^{-1}x))\\ &= -1\cdot(\sin(\cot^{-1}x)\cdot\sin(\cot^{-1}x))\\ &= -\sin^2(\cot^{-1}x)\\ &= -\frac{1}{\csc^2(\cot^{-1}x)}\\ &= -\frac{1}{1+(\cot(\cot^{-1}x))^2}\\ &= -\frac{1}{1+x^2}\\ \end{align} $$

$$ \DeclareMathOperator{\sgn}{sgn} \begin{align} \text{Define } \sgn x&=\begin{cases} 1 & \text{if } x\gt0\\ 0 & \text{if } x=0\\ -1 & \text{if } x\lt0\\ \end{cases}\text{,}\\ \text{then } f(x) &= \sec^{-1}x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{h}\\ &= \lim_{h\to0}\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{x+h-x}\\ &= \lim_{h\to0}\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{\sec(\sec^{-1}(x+h))-\sec(\sec^{-1}x)}\\ &= \lim_{h\to0}\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{\frac{1}{\cos(\sec^{-1}(x+h))}-\frac{1}{\cos(\sec^{-1}x)}}\\ &= \lim_{h\to0}\frac{(\color{red}\sec^{-1}(x+h)-\sec^{-1}x\color{black})\cos(\sec^{-1}(x+h))\cos(\sec^{-1}x)}{\cos(\sec^{-1}x)-\cos(\sec^{-1}(x+h))}\\ &= \lim_{h\to0}\frac{(\color{red}\sec^{-1}(x+h)-\sec^{-1}x\color{black})\cos(\sec^{-1}(x+h))\cos(\sec^{-1}x)}{-2\sin\frac{\color{blue}\sec^{-1}(x+h)+\sec^{-1}x}{2}\sin(-\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x\color{black}}{2})}\\ &=\lim_{\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{2}\to0}\frac{\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{2}}{\sin\frac{\color{red}\sec^{-1}(x+h)-\sec^{-1}x}{2}}\cdot\lim_{h\to0}\frac{\cos(\sec^{-1}(x+h))\cos(\sec^{-1}x)}{\sin\frac{\color{blue}\sec^{-1}(x+h)+\sec^{-1}x}{2}}\\ &= 1\cdot\frac{\cos(\sec^{-1}x)\cdot\cos(\sec^{-1}x)}{\sin\frac{\color{blue}\sec^{-1}x+\sec^{-1}x}{2}}\\ &=\frac{\cos^2(\sec^{-1}x)}{\sin(\sec^{-1}x)}\\ &=\frac{\cos(\sec^{-1}x)}{\tan(\sec^{-1}x)}&(0\leq\sec^{-1}x\leq\frac{\pi}{2}, \tan(\sec^{-1}x)\geq0\text{ when }x\geq0)\\ &=\frac{1}{\sec(\sec^{-1}x)(\sgn x\sqrt{(\sec(\sec^{-1}x))^2-1})}&(\frac{\pi}{2}\lt\sec^{-1}x\lt\pi, \tan(\sec^{-1}x)\lt0\text{ when }x\lt0)\\ &=\sgn x\cdot\frac{1}{x\sqrt{x^2-1}}\\ &=\frac{\sqrt{x^2}}{x}\cdot\frac{1}{x\sqrt{x^2-1}}\\ &=\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\\ \end{align} $$

$$ \DeclareMathOperator{\sgn}{sgn} \begin{align} \text{Define } \sgn x&=\begin{cases} 1 & \text{if } x\gt0\\ 0 & \text{if } x=0\\ -1 & \text{if } x\lt0\\ \end{cases}\text{,}\\ \text{then } f(x) &= \csc^{-1}x\\ \frac{d}{dx}f(x) &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to0}\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{h}\\ &= \lim_{h\to0}\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{x+h-x}\\ &= \lim_{h\to0}\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{\csc(\csc^{-1}(x+h))-\csc(\csc^{-1}x)}\\ &= \lim_{h\to0}\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{\frac{1}{\sin(\csc^{-1}(x+h))}-\frac{1}{\sin(\csc^{-1}x)}}\\ &= \lim_{h\to0}\frac{(\color{red}\csc^{-1}(x+h)-\csc^{-1}x\color{black})\sin(\csc^{-1}(x+h))\sin(\csc^{-1}x)}{\sin(\csc^{-1}x)-\sin(\csc^{-1}(x+h))}\\ &= \lim_{h\to0}\frac{(\color{red}\csc^{-1}(x+h)-\csc^{-1}x\color{black})\sin(\csc^{-1}(x+h))\sin(\csc^{-1}x)}{2\cos\frac{\color{blue}\csc^{-1}(x+h)+\csc^{-1}x}{2}\sin(-\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x\color{black}}{2})}\\ &=-\lim_{\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{2}\to0}\frac{\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{2}}{\sin\frac{\color{red}\csc^{-1}(x+h)-\csc^{-1}x}{2}}\cdot\lim_{h\to0}\frac{\sin(\csc^{-1}(x+h))\sin(\csc^{-1}x)}{\cos\frac{\color{blue}\csc^{-1}(x+h)+\csc^{-1}x}{2}}\\ &=-1\cdot\frac{\sin(\csc^{-1}x)\cdot\sin(\csc^{-1}x)}{\cos\frac{\color{blue}\csc^{-1}x+\csc^{-1}x}{2}}\\ &=-\frac{\sin^2(\csc^{-1}x)}{\cos(\csc^{-1}x)}\\ &=-\frac{\sin(\csc^{-1}x)}{\cot(\csc^{-1}x)}&(0\leq\csc^{-1}x\leq\frac{\pi}{2}, \cot(\csc^{-1}x)\geq0\text{ when }x\geq0)\\ &=-\frac{1}{\csc(\csc^{-1}x)(\sgn x\sqrt{(\csc(\csc^{-1}x))^2-1})}&(-\frac{\pi}{2}\lt\csc^{-1}x\lt0, \cot(\csc^{-1}x)\lt0\text{ when }x\lt0)\\ &=-\sgn x\cdot\frac{1}{x\sqrt{x^2-1}}\\ &=-\frac{\sqrt{x^2}}{x}\cdot\frac{1}{x\sqrt{x^2-1}}\\ &=-\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\\ \end{align} $$