If y-e lnsin -1x-0- x - 1 then dy-dx equals -

The correct option is C 1√(1 x^2) nbsp;y= e ^ln( sin ^ 1 x nbsp;) nbsp; nbsp;=sin ^ 1 x nbsp;⇒ dy / dx = 1 /√( 1 x ^ 2 ) nbsp; ...

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Existence of Limit

If y=e lnsi...

Question

If $y = e^{l n ({sin}^{- 1} x)}, 0 < x \leq 1$ then $\frac{d y}{d x}$ equals -

c o s^{- 1} x

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\frac{1}{{sin}^{- 1} x}

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\frac{1}{\sqrt{1 - x^{2}}}

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\frac{x}{\sqrt{1 - x^{2}}}

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Solution

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Similar questions

^{- 1} (x . \sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^{2}})

\frac{d y}{d x} =

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

0 < x < 1

\sqrt{1 + x^{2}} {[{x c o s ({c o t}^{- 1} x) + s i n ({c o t}^{- 1} x)}^{2} - 1]}^{1 / 2}

\frac{x}{\sqrt{1 + x^{2}}}

x

x \sqrt{1 + x^{2}}

\sqrt{1 + x^{2}}

y = x \sqrt{1 - x^{2}} + s i n^{- 1} x

\frac{d y}{d x}

y = {sin}^{- 1} (x \sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^{2}})

\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}} + p

p

\frac{1}{λ \sqrt{x (1 - x)}}

λ

$F i n d \frac{d y}{d x}, i f y = s i n^{- 1} x + s i n^{- 1} \sqrt{1 - x^{2}}, - 1 \leq x, \leq 1.$

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