# How do you find the derivative of (sqrt(1-x^2))arcsinx?

Sep 7, 2017

$1 - \frac{x \cdot a r c \sin x}{\sqrt{1 - {x}^{2}}} , | x | < 1.$

#### Explanation:

Let, $y = \sqrt{1 - {x}^{2}} \cdot a r c \sin x .$

The desired Derivative can be obtained using the Product Rule for

Diffn.

Here is, an another way to solve the Problem.

We subst. $t = a r c \sin x , | x | \le 1. \Rightarrow \sin t = x , | t | \le \frac{\pi}{2.}$

Hence, $y = \sqrt{1 - {\sin}^{2} t} \cdot t = t \cos t , \text{ where, } t = \sin x \ldots . \left(1\right) .$

Thus, $y \text{ is a fun. of "t," and, t of } x .$

By the Chain Rule, then, we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} ,$

$= \left\{\frac{d}{\mathrm{dt}} \left(t \cos t\right)\right\} \left\{\frac{d}{\mathrm{dx}} \left(a r c \sin x\right)\right\} \ldots \ldots \left[\because , \left(1\right)\right] ,$

$= \left\{\cos t \cdot \frac{d}{\mathrm{dt}} \left(t\right) + t \cdot \frac{d}{\mathrm{dt}} \left(\cos t\right)\right\} \left(\frac{1}{\sqrt{1 - {x}^{2}}}\right) ,$

$= \left\{\cos t + t \left(- \sin t\right)\right\} \cos x ,$

$= \left\{\sqrt{1 - {x}^{2}} - x \cdot a r c \sin x\right\} \left(\frac{1}{\sqrt{1 - {x}^{2}}}\right) ,$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{x \cdot a r c \sin x}{\sqrt{1 - {x}^{2}}} , | x | < 1.$

N.B. : The Domain of $g \text{ is "[-1,1];" and, that of "g'" is } \left(- 1 , 1\right) .$

Enjoy Maths.!