Question

if sec^-1(cosecx)find dy/dx ​

Answer 1

Let y = sec^-1 x

by rewriting in terms of secant

=> sec y = x

by differentiating with respect to x,

=> sec y tan y . y' = 1

by dividing by sec y tan y

=> y' = 1 / sec y tan y

since sec y = x and tan x = √(sec²y - 1) = √(x - 1)

=> y' = 1 / x √(x² - 1)

Answer 2

$\huge \boxed{ \purple{\mathfrak{solution}}}$

Given:

y=

$\boxed {\bold{ {sec}^{ - 1} (cosecx)}}$

To find :

dy/dx =

$\rightarrow \frac{d}{dx} ( {sec}^{ - 1} (cosecx)) \\ \\ \rightarrow \frac{d}{dx} ( {sec}^{ - 1} (sec( \frac{\pi}{2} - x))) \\ \\ \rightarrow \: \frac{d}{dx} ( \frac{\pi}{2} - x) \\ \\ \rightarrow \: \frac{d}{dx} ( \frac{\pi}{2} ) - \frac{d}{dx}( x) \\ \\ \rightarrow \: 0 - 1 \\ \\ \rightarrow \: - 1$

Formula used :

$\boxed{ \bold{ \green{\star \: cosec \: x = sec( \frac{\pi}{2} - x)}}} \\ \\ \boxed{ \bold{ \blue{\star {sec}^{ - 1} (secx) = x}}}$