Question

$\frac{d}{dx}\left[{\sin }^{-1}\frac{x}{a}\right]=$ , $a<0,\left|\frac{x}{a}\right|<1$

• A. $\frac{1}{\sqrt{a^2-x^2}}$
• B. $\frac{1}{\sqrt{x^2-a^2}}$
• C. $-\frac{1}{\sqrt{a^2-x^2}}$
• D. $-\frac{1}{\sqrt{x^2-a^2}}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the value of $\frac{d}{dx}\left[{\sin }^{-1}\frac{x}{a}\right]$:

Let $y={\sin }^{-1}\left(\frac{x}{a}\right)$

Differentiate both sides with respect to $x$

$\frac{dy}{dx}=\frac{d{\sin }^{-1}\left(\frac{x}{a}\right)}{d\left(\frac{x}{a}\right)}\times \frac{d\left(\frac{x}{a}\right)}{dx}$

$\Rightarrow$$\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}\times \frac{1}{a}\times \frac{dx}{dx}$ $\left[\mathbf{\because }\frac{\mathbf{d}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\left(a\right)}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}}}\right]$

$\Rightarrow$$\frac{dy}{dx}=\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\times \frac{1}{a}\times 1$

$\Rightarrow$$\frac{dy}{dx}=\frac{1}{\sqrt{\frac{a^2-x^2}{a^2}}}\times \frac{1}{a}$

$\Rightarrow$$\frac{dy}{dx}=\frac{a}{\sqrt{a^2-x^2}}\times \frac{1}{a}$

$\mathbf{\therefore }\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathbf{x}}^{\mathbf{2}}}}$

Hence, option ‘A’ is Correct.