Derivative examples

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 09 July 2012

Examples

Example: Explain why $\frac{d\ln x}{dx}=\frac{1}{x}.$

Since $e^{\ln x}=x, \frac{d{e}^{\ln x}}{dx}=\frac{dx}{dx}.$
So $e^{\ln x}\frac{d\ln x}{dx}=1.$ So $x\frac{d\ln x}{dx}=1.$ So $\frac{d\ln x}{dx}=\frac{1}{x}.$

Example: Find $\frac{d{\sin }^{-1}x}{dx}.$

Since $\sin \left({\sin }^{-1}x\right)=x, \frac{d\sin \left({\sin }^{-1}x\right)}{dx}=\frac{dx}{dx}.$
So $\cos \left({\sin }^{-1}x\right)\frac{d{\sin }^{-1}x}{dx}=1.$ So $\frac{d{\sin }^{-1}x}{dx}=\frac{1}{\cos \left({\sin }^{-1}x\right)}.$
So we would like to "simplify" $\cos \left({\sin }^{-1}x\right).$
Since $1-{\cos }^2\left({\sin }^{-1}x\right)={\sin }^2\left({\sin }^{-1}x\right), 1-{\left(\cos \left({\sin }^{-1}x\right)\right)}^2={\left(\sin \left({\sin }^{-1}x\right)\right)}^2.$
So $1-{\left(\cos \left({\sin }^{-1}x\right)\right)}^2=x^2.$ So $1-x^2={\left(\cos \left({\sin }^{-1}x\right)\right)}^2.$
So $\cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2}.$ So $\frac{d{\sin }^{-1}x}{dx}=\frac{1}{\cos \left({\sin }^{-1}x\right)}=\frac{1}{\sqrt{1-x^2}}.$

Example: Find $\frac{d{\cos }^{-1}x}{dx}.$

Since $\cos \left({\cos }^{-1}x\right)=x, \frac{d\cos \left({\cos }^{-1}x\right)}{dx}=\frac{dx}{dx}.$
So $-\sin \left({\cos }^{-1}x\right)\frac{d{\cos }^{-1}x}{dx}=1.$ So $\frac{d{\cos }^{-1}x}{dx}=\frac{-1}{\sin \left({\cos }^{-1}x\right)}.$
So we would like to "simplify" $\sin \left({\cos }^{-1}x\right).$
Since $1-{\sin }^2\left({\cos }^{-1}x\right)={\cos }^2\left({\cos }^{-1}x\right), 1-{\left(\sin \left({\cos }^{-1}x\right)\right)}^2={\left(\cos \left({\cos }^{-1}x\right)\right)}^2.$
So $1-{\left(\sin \left({\cos }^{-1}x\right)\right)}^2=x^2.$ So $1-x^2={\left(\sin \left({\cos }^{-1}x\right)\right)}^2.$
So $\sin \left({\cos }^{-1}x\right)=\sqrt{1-x^2}.$ So $\frac{d{\cos }^{-1}x}{dx}=\frac{-1}{\sin \left({\cos }^{-1}x\right)}=\frac{-1}{\sqrt{1-x^2}}.$

Example: Find $\frac{d{\tan }^{-1}x}{dx}.$

Since $\tan \left({\tan }^{-1}x\right)=x, \frac{d\tan \left({\tan }^{-1}x\right)}{dx}=\frac{dx}{dx}.$
So ${\sec }^2\left({\tan }^{-1}x\right)\frac{d{\tan }^{-1}x}{dx}=1.$ So $\frac{d{\tan }^{-1}x}{dx}=\frac{1}{{\sec }^2\left({\tan }^{-1}x\right)}.$
So we would like to "simplify" ${\sec }^2\left({\tan }^{-1}x\right).$
Since ${\sin }^{2}x+{\cos }^{2}x=1,$ $$\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{{\cos }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}.$$ So ${\tan }^{2}x+1={\sec }^{2}x.$
So ${\sec }^2\left({\tan }^{-1}x\right)={\tan }^2\left({\tan }^{-1}x\right)+1={\left(\tan \left({\tan }^{-1}x\right)\right)}^2+1=x^2+1.$
So $\frac{d{\tan }^{-1}x}{dx}=\frac{1}{x^2+1}.$

Example: Find $\frac{d{\cot }^{-1}x}{dx}.$

Since $\cot \left({\cot }^{-1}x\right)=x, \frac{d\cot \left({\cot }^{-1}x\right)}{dx}=\frac{dx}{dx}.$
So $-{\csc }^2\left({\cot }^{-1}x\right)\frac{d{\cot }^{-1}x}{dx}=1.$ So $\frac{d{\cot }^{-1}x}{dx}=\frac{-1}{{\csc }^2\left({\cot }^{-1}x\right)}.$
So we would like to "simplify" ${\csc }^2\left({\cot }^{-1}x\right).$
Since ${\sin }^{2}x+{\cos }^{2}x=1,$ $$\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}=\frac{1}{{\sin }^{2}x}.$$ So $1+{\cot }^{2}x={\csc }^{2}x.$
So ${\csc }^2\left({\cot }^{-1}x\right)=1+{\cot }^2\left({\cot }^{-1}x\right)=1+{\left(\cot \left({\cot }^{-1}x\right)\right)}^2=1+x^2.$
So $\frac{d{\cot }^{-1}x}{dx}=\frac{-1}{1+x^2}.$

Example: Find $\frac{d{\sec }^{-1}x}{dx}.$

Since $\sec \left({\sec }^{-1}x\right)=x, \frac{d\sec \left({\sec }^{-1}x\right)}{dx}=\frac{dx}{dx}.$
So $\tan \left({\sec }^{-1}x\right)\sec \left({\sec }^{-1}x\right)\frac{d{\sec }^{-1}x}{dx}=1.$ So $\tan \left({\sec }^{-1}x\right)\cdot x\cdot \frac{d{\sec }^{-1}x}{dx}=1.$
So $\frac{d{\sec }^{-1}x}{dx}=\frac{1}{x\tan \left({\sec }^{-1}x\right)}.$
So we wouldlike to "simplify" $\tan \left({\sec }^{-1}x\right).$
Since ${\sin }^{2}x+{\cos }^{2}x=1,$ $$\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{{\cos }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}.$$ So ${\tan }^{2}x+1={\sec }^{2}x.$
${\tan }^2\left({\sec }^{-1}x\right)+1={\sec }^2\left({\sec }^{-1}x\right).$ So ${\left(\tan \left({\sec }^{-1}x\right)\right)}^2+1={\left(\sec \left({\sec }^{-1}x\right)\right)}^2.$
So ${\left(\tan \left({\sec }^{-1}x\right)\right)}^2+1=x^2.$ So $\tan \left({\sec }^{-1}x\right)=\sqrt{x^2-1}.$
So $\frac{d{\sec }^{-1}x}{dx}=\frac{1}{x\sqrt{x^2-1}}.$

Example: Find $\frac{d{\csc }^{-1}x}{dx}.$

Since $\csc \left({\csc }^{-1}x\right)=x, \frac{d\csc \left({\csc }^{-1}x\right)}{dx}=\frac{dx}{dx}.$
So $-\csc \left({\csc }^{-1}x\right)\cot \left({\csc }^{-1}x\right)\frac{d{\csc }^{-1}x}{dx}=1.$ So $-x\cot \left({\csc }^{-1}x\right)\frac{d{\csc }^{-1}x}{dx}=1.$
So $\frac{d{\csc }^{-1}x}{dx}=\frac{-1}{x\cot \left({\csc }^{-1}x\right)}.$
So we would like to "simplify" $\cot \left({\csc }^{-1}x\right).$
Since ${\sin }^{2}x+{\cos }^{2}x=1,$ $$\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}=\frac{1}{{\sin }^{2}x}.$$
So $1+{\cot }^{2}x={\csc }^{2}x.$
So $1+{\cot }^2\left({\csc }^{-1}x\right).$ So $1+{\left(\cot \left({\csc }^{-1}x\right)\right)}^2={\left(\csc \left({\csc }^{-1}x\right)\right)}^2.$
So $1+{\left(\cot \left({\csc }^{-1}x\right)\right)}^2=x^2.$ So $\cot \left({\csc }^{-1}x\right)=\sqrt{x^2-1}.$
So $\frac{d{\csc }^{-1}x}{dx}=\frac{-1}{x\sqrt{x^2-1}}.$

Example: Find $\frac{dy}{dx}$ when $y={\log }_{x}10.$

$$x^y=x^{{\log }_{x}10}=10.$$ Take the derivative: $$\begin{array}{rcl}\frac{d{x}^y}{dx}& =& \frac{d{\left(e^{\ln x}\right)}^y}{dx}\\ & =& \frac{d{e}^y\ln x}{dx}\\ & =& e^{y\ln x}(y\cdot \frac{1}{x}+\frac{dy}{dx}\ln x)\\ & =& \frac{d10}{dx}=0.\end{array}$$ So $e^{y\ln x}(y\cdot \frac{1}{x}+\frac{dy}{dx}\ln x)=0.$
Solve for $\frac{dy}{dx}.$ $$e^{y\ln x}\frac{dy}{dx}\ln x=\frac{-e^{y\ln x}y}{x}.$$ So $\frac{dy}{dx}=\frac{-e^{y\ln x}y}{x{e}^{y\ln x}\ln x}=\frac{-y}{x\ln x}=\frac{{\log }_{x}10}{x\ln x}.$

Example: Find the third derivative of $2^x$ with respect to $x.$

$$\begin{array}{rcl}y& =& 2^x.\\ \frac{dy}{dx}& =& \frac{d{2}^x}{dx}=\frac{d{\left(e^{\ln 2}\right)}^x}{dx}=\frac{d{e}^{x\ln 2}}{dx}=e^{x\ln 2}\left(\ln 2\right)={\left(e^{\ln 2}\right)}^x\ln 2=2^x\ln 2.\\ y& =& \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d{2}^x\ln 2}{dx}=\ln 2\cdot 2^x\ln 2={\left(\ln 2\right)}^2{2}^x.\\ \frac{d^{3}y}{d{x}^3}& =& \frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=\frac{d}{dx}\left({\left(\ln 2\right)}^2{2}^x\right)={\left(\ln 2\right)}^2{2}^x\ln 2={\left(\ln 2\right)}^3{2}^x.\end{array}$$

Example: If $y=a\cos \left(\ln x\right)+b\sin \left(\ln x\right)$ show that $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0.$

$$\begin{array}{rcl}\frac{dy}{dx}& =& a(-\sin \left(\ln x\right))\frac{1}{x}+b\cos \left(\ln x\right)\frac{1}{x}\\ & =& -a\sin \left(\ln x\right)x^{-1}+b\cos \left(\ln x\right)x^{-1},\\ & & \\ \frac{d^{2}y}{d{x}^2}& =& -a\cos \left(\ln x\right)\frac{1}{x}{x}^{-1}+(-a)\sin \left(\ln x\right)(-1)x^{-2}+(-b)\sin \left(\ln x\right)\frac{1}{x}{x}^{-1}+b\cos \left(\ln x\right)(-1)x^{-2}\\ & =& \frac{-a\cos \left(\ln x\right)+a\sin \left(\ln x\right)-b\sin \left(\ln x\right)-b\cos \left(\ln x\right)}{x^2}\\ & =& \frac{1}{x^2}((a-b)\sin \left(\ln x\right)-(a+b)\cos \left(\ln x\right)).\end{array}$$ So $$\begin{array}{rcl}LHS& =& x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y\\ & =& x^2\frac{1}{x^2}((a-b)\sin \left(\ln x\right)-(a+b)\cos \left(\ln x\right))\\ & & +x(-a\sin \left(\ln x\right)x^{-1}+b\cos \left(\ln x\right)x^{-1})\\ & & +a\cos \left(\ln x\right)+b\sin \left(\ln x\right)\\ & =& (a-b)\sin \left(\ln x\right)-(a+b)\cos \left(\ln x\right)\\ & & -a\sin \left(\ln x\right)+b\cos \left(\ln x\right)\\ & & +b\sin \left(\ln x\right)+a\cos \left(\ln x\right)\\ & =& 0.\end{array}$$

Example: Find $\frac{dy}{dx}$ when $a\sin \left(xy\right)+b\cos \left(\frac{x}{y}\right)=0.$

Take the derivative: $$\begin{array}{rcl}0& =& a\cos \left(xy\right)(x\frac{dy}{dx}+1\cdot y)+(-b)\sin \left(\frac{x}{y}\right)(x(-1)y^{-2}\frac{dy}{dx}+1\cdot y^{-1})\\ & =& a\cos \left(xy\right)x\frac{dy}{dx}+a\cos \left(xy\right)y+b\sin \left(\frac{x}{y}\right)\frac{x}{y^2}\frac{dy}{dx}-b\sin \left(\frac{x}{y}\right)y^{-1}.\end{array}$$ Solve for $\frac{dy}{dx}.$ $$a\cos \left(xy\right)x\frac{dy}{dx}+b\sin \left(\frac{x}{y}\right)\frac{x}{y^2}\frac{dy}{dx}=a\cos \left(xy\right)y-b\sin \left(\frac{x}{y}\right)y^{-1}.$$ So $$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{a\cos \left(xy\right)y-b\sin \left(\frac{x}{y}\right)y^{-1}}{a\cos \left(xy\right)x+b\sin \left(\frac{x}{y}\right)\frac{x}{y^2}}\\ & =& \frac{a\cos \left(xy\right)y^3-b\sin \left(\frac{x}{y}\right)y}{a\cos \left(xy\right)x{y}^2+b\sin \left(\frac{x}{y}\right)x}.\end{array}$$

Example: Find $\frac{dy}{dx}$ when $y={\tan }^{-1}\left(\frac{a}{x}\right)\cdot {\cot }^{-1}\left(\frac{x}{a}\right).$

$$\begin{array}{rcl}\frac{dy}{dx}& =& {\tan }^{-1}\left(\frac{a}{x}\right)\left(\frac{-1}{1+{\left(\frac{x}{a}\right)}^2}\right)\frac{1}{a}+\frac{1}{1+{\left(\frac{x}{a}\right)}^2}(-1)a{x}^{-2}{\cot }^{-1}\left(\frac{x}{a}\right)\\ & =& \frac{-{\tan }^{-1}\left(\frac{a}{x}\right)}{a+\frac{x^2}{a}}+\frac{-{\cot }^{-1}\left(\frac{x}{a}\right)a}{x^2+a^2}\\ & =& \frac{-{\tan }^{-1}\left(\frac{a}{x}\right)a}{a^2+x^2}+\frac{-{\cot }^{-1}\left(\frac{x}{a}\right)a}{x^2+a^2}\\ & =& \left(\frac{-a}{a^2+x^2}\right)({\tan }^{-1}\left(\frac{a}{x}\right)+{\cot }^{-1}\left(\frac{x}{a}\right)).\end{array}$$ If $\frac{a}{x}=\tan z$ then $\frac{x}{a}=\cot z$ and $z={\tan }^{-1}\left(\frac{a}{x}\right)={\cot }^{-1}\left(\frac{x}{a}\right).$ So $$\frac{dy}{dx}=\left(\frac{-a}{a^2+x^2}\right)({\tan }^{-1}\left(\frac{a}{x}\right)+{\tan }^{-1}\left(\frac{a}{x}\right))=\frac{-2a{\tan }^{-1}\left(\frac{a}{x}\right)}{a^2+x^2}.$$

Example: Find $\frac{d{e}^x}{dx}.$

$$\begin{array}{rcl}\frac{d{e}^x}{dx}& =& \frac{d}{dx}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\cdots )\\ & =& 0+1+\frac{1}{2!}2x+\frac{1}{3!}3x^2+\frac{1}{4!}4x^3+\frac{1}{5!}5x^4+\frac{1}{6!}6x^5+\frac{1}{7!}7x^6+\cdots \\ & =& 1+\frac{1}{2}2x+\frac{1}{3\cdot 2!}3x^2+\frac{1}{4\cdot 3!}4x^3+\frac{1}{5\cdot 4!}5x^4+\frac{1}{6\cdot 5!}6x^5+\frac{1}{7\cdot 6!}7x^6+\cdots \\ & =& 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\cdots \\ & =& e^x.\end{array}$$

Example: Find $\frac{d\sin x}{dx}.$

$$\begin{array}{rcl}\frac{d\sin x}{dx}& =& \frac{d}{dx}(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}-\cdots )\\ & =& 1-\frac{1}{3!}3x^2+\frac{1}{5!}5x^4-\frac{1}{7!}7x^6+\frac{1}{9!}9x^8-\frac{1}{11!}11x^{10}+\frac{1}{13!}13x^{12}-\cdots \\ & =& 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+\frac{x^{12}}{12!}-\cdots \\ & =& \cos x.\end{array}$$

Example: Find $\frac{d\cos x}{dx}.$

$$\begin{array}{rcl}\frac{d\cos x}{dx}& =& \frac{d}{dx}(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+\frac{x^{12}}{12!}-\cdots )\\ & =& 0-\frac{1}{2!}2x+\frac{1}{4!}4x^3-\frac{1}{6!}6x^5+\frac{1}{8!}8x^7-\frac{1}{10!}10x^9+\frac{1}{12!}12x^{11}-\cdots \\ & =& -x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-\frac{x^9}{9!}+\frac{x^{11}}{11!}-\cdots \\ & =& -(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\cdots )\\ & =& -\sin x.\end{array}$$

Example: Find $\frac{d\tan x}{dx}.$

$$\begin{array}{rcl}\frac{d\tan x}{dx}& =& \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)\\ & =& \frac{d}{dx}\left(\sin x{\left(\cos x\right)}^{-1}\right)\\ & =& \sin x\frac{d{\left(\cos x\right)}^{-1}}{dx}+\frac{d\sin x}{dx}{\left(\cos x\right)}^{-1}\\ & =& \sin x(-1){\left(\cos x\right)}^{-2}\frac{d\cos x}{dx}+\cos x\cdot \frac{1}{\cos x}\\ & =& -\frac{\sin x}{{\cos }^{2}x}(-\sin x)+1\\ & =& \frac{{\sin }^{2}x}{{\cos }^{2}x}+1\\ & =& \frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x}\\ & =& \frac{1}{{\cos }^{2}x}\\ & =& {\sec }^{2}x.\end{array}$$

Example: Find $\frac{d\sec x}{dx}.$

$$\begin{array}{rcl}\frac{d\sec x}{dx}& =& \frac{d}{dx}\left(\frac{1}{\cos x}\right)\\ & =& \frac{d}{dx}\left({\left(\cos x\right)}^{-1}\right)\\ & =& (-1){\left(\cos x\right)}^{-2}\frac{d\cos x}{dx}\\ & =& -\frac{1}{{\cos }^{2}x}(-\sin x)\\ & =& \frac{\sin x}{{\cos }^{2}x}\\ & =& \frac{\sin x}{\cos x}\cdot \frac{1}{\cos x}\\ & =& \tan x\sec x.\end{array}$$

Example: Find $\frac{d\csc x}{dx}.$

$$\begin{array}{rcl}\frac{d\csc x}{dx}& =& \frac{d}{dx}\left(\frac{1}{\sin x}\right)\\ & =& \frac{d}{dx}\left({\left(\sin x\right)}^{-1}\right)\\ & =& (-1){\left(\sin x\right)}^{-2}\frac{d\sin x}{dx}\\ & =& -\frac{1}{{\sin }^{2}x}\left(\cos x\right)\\ & =& -\frac{\cos x}{{\sin }^{2}x}\\ & =& -\frac{\cos x}{\sin x}\cdot \frac{1}{\sin x}\\ & =& -\cot x\csc x.\end{array}$$

Example: Find $\frac{d\cot x}{dx}.$

$$\begin{array}{rcl}\frac{d\cot x}{dx}& =& \frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)\\ & =& \frac{d}{dx}\left(\cos x{\left(\sin x\right)}^{-1}\right)\\ & =& \cos x\frac{d{\left(\sin x\right)}^{-1}}{dx}+\frac{d\cos x}{dx}{\left(\sin x\right)}^{-1}\\ & =& \cos x(-1){\left(\sin x\right)}^{-2}\frac{d\sin x}{dx}+(-\sin x)\cdot \frac{1}{\sin x}\\ & =& -\frac{\cos x}{{\sin }^{2}x}\cdot \cos x-1\\ & =& \frac{-{\cos }^{2}x}{{\sin }^{2}x}-1\\ & =& \frac{-{\cos }^{2}x-{\sin }^{2}x}{{\sin }^{2}x}\\ & =& \frac{-1}{{\sin }^{2}x}\\ & =& {\csc }^{2}x.\end{array}$$

Example: Find $\frac{d\sinh x}{dx}.$

$$\begin{array}{rcl}\frac{d\sinh x}{dx}& =& \frac{d}{dx}(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\frac{x^9}{9!}+\frac{x^{11}}{11!}+\frac{x^{13}}{13!}+\cdots )\\ & =& 1+\frac{1}{3!}3x^2+\frac{1}{5!}5x^4+\frac{1}{7!}7x^6+\frac{1}{9!}9x^8+\frac{1}{11!}11x^{10}+\frac{1}{13!}13x^{12}+\cdots \\ & =& 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+\frac{x^{10}}{10!}+\frac{x^{12}}{12!}+\cdots \\ & =& \cosh x.\end{array}$$

Example: Find $\frac{d\cosh x}{dx}.$

$$\begin{array}{rcl}\frac{d\cosh x}{dx}& =& \frac{d}{dx}(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+\frac{x^{10}}{10!}+\frac{x^{12}}{12!}+\cdots )\\ & =& 0+\frac{1}{2!}2x+\frac{1}{4!}4x^3+\frac{1}{6!}6x^5+\frac{1}{8!}8x^7+\frac{1}{10!}10x^9+\frac{1}{12!}12x^{11}+\cdots \\ & =& x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\frac{x^9}{9!}+\frac{x^{11}}{11!}+\cdots \\ & =& \sinh x.\end{array}$$

Example: Find $\frac{d\tanh x}{dx}.$

$$\begin{array}{rcl}\frac{d\tanh x}{dx}& =& \frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right)\\ & =& \frac{d}{dx}\left(\sinh x{\left(\cosh x\right)}^{-1}\right)\\ & =& \sinh x\frac{d{\left(\cosh x\right)}^{-1}}{dx}+\frac{d\sinh x}{dx}{\left(\cosh x\right)}^{-1}\\ & =& \sinh x(-1){\left(\cosh x\right)}^{-2}\frac{d\cosh x}{dx}+\cosh x\cdot \frac{1}{\cosh x}\\ & =& -\frac{\sinh x}{{\cosh }^{2}x}\cdot \sinh x+1\\ & =& -\frac{{\sinh }^{2}x}{{\cosh }^{2}x}+1\\ & =& \frac{-{\sinh }^{2}x+{\cosh }^{2}x}{{\cosh }^{2}x}\\ & =& \frac{1}{{\cosh }^{2}x}\\ & =& {\mathrm{sech}}^{2}x.\end{array}$$

Example: Find $\frac{d\mathrm{sech}x}{dx}.$

$$\begin{array}{rcl}\frac{d\mathrm{sech}x}{dx}& =& \frac{d}{dx}\left(\frac{1}{\cosh x}\right)\\ & =& \frac{d}{dx}\left({\left(\cosh x\right)}^{-1}\right)\\ & =& (-1){\left(\cosh x\right)}^{-2}\frac{d\cosh x}{dx}\\ & =& -\frac{1}{{\cosh }^{2}x}\cdot \sinh x\\ & =& -\frac{\sinh x}{{\cosh }^{2}x}\\ & =& -\frac{\sinh x}{\cosh x}\cdot \frac{1}{\cosh x}\\ & =& -\tanh x\mathrm{sech}x.\end{array}$$

Example: Find $\frac{d\mathrm{csch}x}{dx}.$

$$\begin{array}{rcl}\frac{d\mathrm{csch}x}{dx}& =& \frac{d}{dx}\left(\frac{1}{\sinh x}\right)\\ & =& \frac{d}{dx}\left({\left(\sinh x\right)}^{-1}\right)\\ & =& (-1){\left(\sinh x\right)}^{-2}\frac{d\sinh x}{dx}\\ & =& -\frac{1}{{\sinh }^{2}x}\left(\cosh x\right)\\ & =& -\frac{\cosh x}{{\sinh }^{2}x}\\ & =& -\frac{\cosh x}{\sinh x}\cdot \frac{1}{\sinh x}\\ & =& -\coth x\mathrm{csch}x.\end{array}$$

Example: Find $\frac{d\coth x}{dx}.$

$$\begin{array}{rcl}\frac{d\coth x}{dx}& =& \frac{d}{dx}\left(\frac{1}{\tanh x}\right)\\ & =& \frac{d{\left(\tanh x\right)}^{-1}}{dx}\\ & =& (-1){\left(\tanh x\right)}^{-2}\frac{d\tanh x}{dx}\\ & =& -\frac{1}{{\tanh }^{2}x}\frac{d\tanh x}{dx}\\ & =& -\frac{1}{{\tanh }^{2}x}\cdot {\mathrm{sech}}^{2}x\\ & =& -\frac{1}{\frac{{\sinh }^{2}x}{{\cosh }^{2}x}}\cdot \frac{1}{{\cosh }^{2}x}\\ & =& -\frac{1}{{\sinh }^{2}x}\\ & =& -{\mathrm{csch}}^{2}x.\end{array}$$

References

References?

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