Y-log tanx2 - sin-1cos x- then dydx is

The correct option is A cosec (x) 1 dy/dx=1/tan x2^2x2·1/2+1/√(1 cos^2x)( sin x) =1/2sinx2·cosx2 1 =cosec (x) 1 ...

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Graphs of Basic Inverse Trigonometric Functions

y=log tanx2 +...

Question

${y = log (tan \frac{x}{2}) + sin}^{- 1} (cos x)$ , then $\frac{d y}{d x}$ is

cosec (x) - 1

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cosec x

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cosec (x) + 1

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x

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Similar questions

\int \frac{cosec x}{{cos}^{2} (1 + log tan \frac{x}{2})} d x

x,

cosec x - cot x = \frac{1}{(cosec x + cot x)} .

y = log tan \frac{x}{2} + {sin}^{- 1} (cos x), \frac{d y}{d x} =

List I	List II
A) $\frac{d y}{d x} - y cot x = csc x$	1. $x sin x$
B) $x log x \frac{d y}{d x} + y = 2 log x$	2. $x e^{x}$
C) $x sin x \frac{d y}{d x} + y (x cos x + sin x) = sin x$	3. $cos e c x$
D) $x \frac{d y}{d x} + y (1 + x) = 1$	4. $- cos e c x$
	5. $l o g x$

\frac{d y}{d x}

y = log (tan \frac{x}{2}) + {sin}^{- 1} (cos x)

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