The correct option is A 1/√( ( 1 x^2 )) #160;y= sin ^ 1x, y+δ y= sin ^ 1 ( x+δ x ) #160;∴ ( x+δ x ) = sin ( y+δ y ) #160;dy/dx= lim_ ...

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Evaluation of a Determinant

dsin-1x/dx

Question

$\frac{d (s i n^{- 1} x)}{d x}$

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

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\frac{1}{\sqrt{(1 - x^{4})}}

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\frac{1}{\sqrt{(1 - x^{3})}}

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

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Solution

The correct option is A $\frac{1}{\sqrt{(1 - x^{2})}}$
$y = {sin}^{- 1} x,$
$y + δ y = {sin}^{- 1} (x + δ x)$
$∴ (x + δ x) = sin (y + δ y)$
$\frac{d y}{d x} = lim x \to 0 \frac{1}{δ x} ({sin}^{- 1} (x + δ x) - {sin}^{- 1} x)$
$= lim x \to 0 \frac{δ y}{sin (y + δ y) - sin y} = \frac{1}{cos y}$
$= \frac{1}{\sqrt{(1 - {sin}^{2} y)}} = \frac{1}{\sqrt{(1 - x^{2})}}$
Note that when $: δ x \to 0, δ y$ also tends to zero.

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d(sin−1x)dx

The correct option is A 1√(1−x2)y=sin−1x,y+δy=sin−1(x+δx)∴(x+δx)=sin(y+δy)dydx=limx→01δx(sin−1(x+δx)−sin−1x)=limx→0δysin(y+δy)−siny=1cosy=1√(1−sin2y)=1√(1−x2) Note that when :δx→0,δy also tends to zero.

$\frac{d (s i n^{- 1} x)}{d x}$