If y-1-cos 2 x 1-cos 2 x- x - 0- - 2 - 2- - then dyd x is equal to

We have, y=1 cos nbsp;2x1+cos nbsp;2x, nbsp;x∈0,π2∪π2,π⇒y=2 nbsp;sin2 nbsp;x2 nbsp;cos2 nbsp;x=tan2 nbsp;x⇒y=tan nbsp;x, nbsp;where nbsp; ...

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Derivative

If y=1-cos 2 ...

Question

If $y = \sqrt{\frac{1 - cos 2 x}{1 + cos 2 x}}, x \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π)$ , then $\frac{d y}{d x}$ is equal to

s e c^{2} x, i f x \in (0, \frac{π}{2})

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- s e c^{2} x, i f x \in (\frac{π}{2}, π)

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s e c^{2} x, i f x \in (\frac{π}{2}, π)

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- s e c^{2} x, i f x \in (0, \frac{π}{2})

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Solution

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Similar questions

\frac{d y}{d x} at x = \sqrt{\frac{π}{2}}

y = \sin^{- 1} (\sin x), - \frac{π}{2} \leq x \leq \frac{π}{2}

\frac{d y}{d x} for x \in (- \frac{π}{2}, \frac{π}{2}) .

π < x < \frac{3 π}{2}

\sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}}

\leq x

∣ ∣ ∣ ∣ ∣ \begin{matrix} [e] & [π] & [π^{2} - 6] [π] & [π^{2} - 6] & [e] [π^{2} - 6] & [e] & [π] \end{matrix} ∣ ∣ ∣ ∣ ∣

[x]

\leq x

∣ ∣ ∣ ∣ ∣ \begin{matrix} [e] & [π] & [π^{2} - 6] [π] & π^{2} - 6 & [e] [π^{2} - 6] & [e] & [π] \end{matrix} ∣ ∣ ∣ ∣ ∣

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