Derivative of Arccos x

What is the Derivative of Arccos x ?

The derivative of arccos x is -1/√1-x².

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

$\frac{d}{dx}(arccosx)=-\frac{1}{\sqrt{1-x^2}}$

Proof of the Derivative of Arccos x

Way 1

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

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

$arccos(x+h)=U$, $cosU=x+h$

$arccosx=V$, $cosV=x$

$cosU-cosV=\cancel{x}+h-\cancel{x}$

$cosU-cosV=h$

$h\to 0$, $cosU-cosV\to 0$, $cosU\to cosV$, $U\to V$

$(arccosx{)}^{\prime }=\underset{U\to V}{\lim }\frac{U-V}{cosU-cosV}$

$\boxed{cosp-cosq=-2.sin\frac{p+q}{2}.sin\frac{p-q}{2}}$

$(arccosx{)}^{\prime }=\underset{U\to V}{\lim }\frac{U-V}{-2.sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arccosx{)}^{\prime }=\underset{U\to V}{\lim }\frac{-\frac{1}{2}.(U-V)}{sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arccosx{)}^{\prime }=\underset{U\to V}{\lim }\frac{-\frac{U-V}{2}}{sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arccosx{)}^{\prime }=-\underset{U\to V}{\lim }\frac{\frac{U-V}{2}}{sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arccosx{)}^{\prime }=-\underset{U\to V}{\lim }(\frac{\frac{U-V}{2}}{sin\frac{U-V}{2}}.\frac{1}{sin\frac{U+V}{2}})$

$(arccosx{)}^{\prime }=-\underset{U\to V}{\lim }\frac{\frac{U-V}{2}}{sin\frac{U-V}{2}}.\underset{U\to V}{\lim }\frac{1}{sin\frac{U+V}{2}}$

$U\to V$, $\frac{U-V}{2}\to 0$

$(arccosx{)}^{\prime }=-\underset{\frac{U-V}{2}\to 0}{\lim }\frac{\frac{U-V}{2}}{sin\frac{U-V}{2}}.\underset{U\to V}{\lim }\frac{1}{sin\frac{U+V}{2}}$

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

$(arccosx{)}^{\prime }=-1.\frac{1}{sin\frac{V+V}{2}}$

$(arccosx{)}^{\prime }=-\frac{1}{sin\frac{V+V}{2}}$

$(arccosx{)}^{\prime }=-\frac{1}{sin\frac{\cancel{2}V}{\cancel{2}}}$

$(arccosx{)}^{\prime }=-\frac{1}{sinV}$

$(arccosx{)}^{\prime }=-\frac{1}{sinV}$

$cosV=x$

$sinV=\sqrt{1-x^2}$

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

Way 2

$y=arccosx$

$cosy=x$

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

$\boxed{(cosu{)}^{\prime }=-u^{\prime }.sinu}$

$-y^{\prime }.siny=1$

$y^{\prime }=-\frac{1}{siny}$

$cosy=x$

$siny=\sqrt{1-x^2}$

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