Question

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

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

Answer 1

Answer: (C)

Let, $y={\cos }^{-1}\left(\frac{4x^3}{27}-x\right)$
$={\cos }^{-1}\left(4{\left(\frac{x}{3}\right)}^3-3\left(\frac{x}{3}\right)\right)$
Let $\frac{x}{3}=\cos A$ then, $A={\cos }^{-1}\frac{x}{3}$
Also, $y={\cos }^{-1}\left(4{\cos }^{3}A-3\cos A\right)$
$={\cos }^{-1}(\cos 3A)$
$=3A=3{\cos }^{-1}\left(\frac{x}{3}\right)$
$\therefore \frac{dy}{dx}=\frac{d}{dx}3{\cos }^{-1}\left(\frac{x}{3}\right)$
$=3\times \frac{-1}{\sqrt{1-{\left(\frac{x}{3}\right)}^2}}\times \frac{d}{dx}\left(\frac{x}{3}\right)$
$=3\times \frac{-3}{\sqrt{9-x^2}}\times \frac{1}{3}=-\frac{3}{\sqrt{9-x^2}}$