Question

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

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