Question

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

$\frac{2}{1 + x^{2}}$
$0$
$2$
$1$

Answer: 0

Solution :-

We know that {sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2}

Now, Differentiate with respect to x, we get

\begin{aligned} \frac{d}{d x} ({sin}^{- 1} x + {cos}^{- 1} x) & = & \frac{d (\frac{π}{2})}{d x} \\ ∴ \frac{d}{d x} ({sin}^{- 1} x + {cos}^{- 1} x) & = & 0 \cdot (∵ \frac{d (c o n s t a n t)}{d x} = 0) \end{aligned}

So, correct answer is option-2, $\to 0$

#Differentiation #BSEB PYQ 2017 #Mathematics #BSEB PYQs