Question

$\frac{d}{dx}\left({\cos }^{-1}(\frac{4x^3}{27}-x)\right)=?$

• A. $\frac{3}{\sqrt{x^2-9}}$
• B. $\frac{1}{\sqrt{9-x^2}}$
• C. $\frac{-3}{\sqrt{9-x^2}}$
• D. $\frac{-1}{\sqrt{9-x^2}}$

Answer 1

The correct option is A $\frac{3}{\sqrt{x^2-9}}$

$\frac{d}{dx}\left({\cos }^{-1}(\frac{4\times 3}{27}-x)\right)=\left(\frac{-1}{\sqrt{1-u^2}}\right)(\frac{4\times 2}{9}-1)$

$u=\frac{4\times 3}{27}-x$

$=\frac{3(4x^2-9)}{\sqrt{-{(9-4x^2)}^2(x^2-9)}}=\frac{3(4x^2-9)}{(4x^2-9)\sqrt{(x^2-9)}}$

$=\frac{3}{\sqrt{x^2-9}}$