Question

$\frac{d}{d x} ({sin}^{- 1} x) =$

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

Answer: $\frac{1}{\sqrt{1 - x^{2}}}$

Solution :-

Let {sin}^{- 1} x = y ⟹ x = sin y \dots \dots (1)

Now, Differentiate with respect to x, we get

\begin{aligned} \frac{d (sin y)}{d x} & = & \frac{d (x)}{d x} \\ \frac{d (sin y)}{d y} \cdot \frac{d (y)}{d x} & = & 1 \\ cos y \cdot \frac{d y}{d x} & = & 1 (∵ \frac{d (sin x)}{d x} = cos x) \\ \frac{d y}{d x} & = & \frac{1}{cos y} \\ \frac{d y}{d x} & = & \frac{1}{\sqrt{1 - {sin}^{2} y}} \\ \frac{d y}{d x} & = & \frac{1}{\sqrt{1 - x^{2}}} (∵ sin y = x from eq(1)) \\ ∴ \frac{d ({sin}^{- 1} x)}{d x} & = & \frac{1}{\sqrt{1 - x^{2}}} \cdot \end{aligned}

So, correct answer is option-3, $\to \frac{1}{\sqrt{1 - x^{2}}}$

#Differentiation #BSEB PYQ 2017 #Mathematics #BSEB PYQs