Question

if y=cosec-¹(1/2x√1-x²) ,0<x<1/√2, then find dy/dx​

Answer 1

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Answer 2

Given:

y = $cosec^{-1}$(1/2x√1-x²)

0<x<1/√2

To Find:

dy/dx

Solution:

⇒ y - $cosec^{-1}(\frac{1}{2x\sqrt{1-x^{2} } })$

x = sinθ     θ = $sin^{-1}x$

y = $cosec^{-1}$(1/2sinθcosθ)

y = $cosec^{-1}$(1/sin²θ)                    [sin²θ = 2sinθcosθ]

y = $cosec^{-1}$(cosec2θ)

y = 2θ

y = 2$sin^{-1}x$

So, dy/dx = 2/$\sqrt{1-x^{2} }$