Question

Evaluate:

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

• A. $\frac{4}{\sqrt{9-x^2}}$
• B. $\frac{1}{\sqrt{9-x^2}}$
• C. $\frac{-3}{\sqrt{9-x^2}}$
• D. $\frac{-1}{\sqrt{9-x^2}}$

Answer 1

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

Given,

$\arccos (\frac{4x^3}{27}-x)$

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

$f=\arccos \left(u\right),u=\frac{4x^3}{27}-x$

$=\frac{d}{du}\left(\arccos \left(u\right)\right)\frac{d}{dx}(\frac{4x^3}{27}-x)$

$=(-\frac{1}{\sqrt{1-u^2}})(\frac{4x^2}{9}-1)$

$=(-\frac{1}{\sqrt{1-{(\frac{4x^3}{27}-x)}^2}})(\frac{4x^2}{9}-1)$

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

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