Question

differentiation of cot inverse x

Answer 1

Differentiation of
$\cot^{ - 1}x \: is \: - - - - -$
- cosec x cot x



Hope it will help you .

Answer 2

Answer:

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

Step-by-step explanation:

Formula used:

$1.tan^{-1}x+cot^{-1}x=\frac{\pi}{2}$

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

Now,

$tan^{-1}x+cot^{-1}x=\frac{\pi}{2}$

Differentiate with respect to 'x'

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

$\frac{1}{1+x^2}+\frac{d(cot^{-1}x)}{dx}=0$

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