Derivative of Arccot x

What is the Derivative of Arccot x ?

The derivative of arccot x is -1/1+x².

$(arccotx{)}^{\prime }=-\frac{1}{1+x^2}$

$\frac{d}{dx}(arccotx)=-\frac{1}{1+x^2}$

Proof of the Derivative of Arccot x

Way 1

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(arccotx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{arccot(x+h)-arccotx}{h}$

$arccot(x+h)=U$, $cotU=x+h$

$arccotx=V$, $cotV=x$

$cotU-cotV=\cancel{x}+h-\cancel{x}$

$cotU-cotV=h$

$h\to 0$, $cotU-cotV\to 0$, $cotU\to cotV$, $U\to V$

$(arccotx{)}^{\prime }=\underset{U\to V}{\lim }\frac{U-V}{cotU-cotV}$

$\boxed{cotp-cotq=-\frac{sin(p-q)}{sinp.sinq}}$

$(arccotx{)}^{\prime }=\underset{U\to V}{\lim }\frac{U-V}{-\frac{sin(U-V)}{sinU.sinV}}$

$(arccotx{)}^{\prime }=\underset{U\to V}{\lim }[-\frac{(U-V).sinU.sinV}{sin(U-V)}]$

$(arccotx{)}^{\prime }=-\underset{U\to V}{\lim }[\frac{(U-V).sinU.sinV}{sin(U-V)}]$

$(arccotx{)}^{\prime }=-\underset{U\to V}{\lim }[\frac{U-V}{sin(U-V)}.sinU.sinV]$

$(arccotx{)}^{\prime }=-\underset{U\to V}{\lim }\frac{U-V}{sin(U-V)}.\underset{U\to V}{\lim }sinU.\underset{U\to V}{\lim }sinV$

$U\to V$, $U-V\to 0$

$(arccotx{)}^{\prime }=-\underset{U-V\to 0}{\lim }\frac{U-V}{sin(U-V)}.\underset{U\to V}{\lim }sinU.\underset{U\to V}{\lim }sinV$

$\boxed{\underset{t\to 0}{\lim }\frac{t}{sint}=1}$

$(arccotx{)}^{\prime }=-1.sinV.sinV$

$(arccotx{)}^{\prime }=-si{n}^{2}V$

$cotV=x$

$sinV=\frac{1}{\sqrt{1+x^2}}$

$(arccotx{)}^{\prime }=-(\frac{1}{\sqrt{1+x^2}}{)}^2$

$(arccotx{)}^{\prime }=-\frac{1}{1+x^2}$

Way 2

$y=arccotx$

$coty=x$

$(coty{)}^{\prime }=(x{)}^{\prime }$

$\boxed{(cotu{)}^{\prime }=-u^{\prime }.(1+co{t}^{2}u)}$

$-y^{\prime }.(1+co{t}^{2}y)=1$

$y^{\prime }=-\frac{1}{1+co{t}^{2}y}$

$y^{\prime }=-\frac{1}{1+x^2}$