Question

$\frac{d}{dx}\{{\cos }^{-1}x+{\sin }^{-1}\sqrt{1-x^2}\}=$

• A. $0$
• B. $1$
• C. $\frac{2}{\sqrt{1-x^2}}$
• D. $\frac{-2}{\sqrt{1-x^2}}$

Answer 1

Answer: (D)

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

$\frac{d}{dx}[co{s}^{-1}x+si{n}^{-1}\sqrt{1+x^2}]$

$=\frac{d}{dx}co{s}^{-1}x+\frac{d}{dx}.si{n}^{-1}\sqrt{1-x^2}$

$=\frac{-1}{\sqrt{1-x^2}}+\frac{1}{1-\sqrt{(1-x^2)}}.\frac{d}{dx}\sqrt{1-x^2}$

$=\frac{-1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{x^2}}-\frac{1}{2\sqrt{1-x^2}}-\frac{d}{dx}(1-x^2)$

$=\frac{-1}{\sqrt{1-x^2}}+\frac{1}{x}.\frac{1}{2\sqrt{1-x^2}}.(0-2x)$

$=\frac{-2}{1-x^2}$