\(\frac{d}{d x}\left(\cos ^{-1}\left(\frac{4 x^3}{27}-x\right)\right)=\)
\(\frac{d}{d x}\left(\cos ^{-1}\left(\frac{4 x^3}{27}-x\right)\right)=\)
- \(\frac{3}{\sqrt{9-x^2}}\)
- \(\frac{1}{\sqrt{9-x^2}}\)
- \(\frac{-3}{\sqrt{9-x^2}}\)
- \(\frac{-1}{\sqrt{9-x^2}}\)
Solution
Let,
\(\begin{aligned}
& y=\cos ^{-1}\left(\frac{4 x^3}{27}-x\right) \\
& =\cos ^{-1}\left(4\left(\frac{x}{3}\right)^3-3\left(\frac{x}{3}\right)\right)
\end{aligned}\)
Let \(\quad \frac{x}{3}=\cos A\) then, \(A=\cos ^{-1} \frac{x}{3}\)
Also,
\(\begin{aligned}
y & =\cos ^{-1}\left(4 \cos ^3 A-3 \cos A\right) \\
& =\cos ^{-1}(\cos 3 A) \\
& =3 A=3 \cos ^{-1}\left(\frac{x}{3}\right)
\end{aligned}\)
\(\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{d}{d x} 3 \cos ^{-1}\left(\frac{x}{3}\right) \\
& =3 \times \frac{-1}{\sqrt{1-\left(\frac{x}{3}\right)^2}} \times \frac{d}{d x}\left(\frac{x}{3}\right) \\
& =3 \times \frac{-3}{\sqrt{9-x^2}} \times \frac{1}{3}=-3 / \sqrt{9-x^2}
\end{aligned}\)
Asked in: AP EAMCET 2020 (17 Sep Shift 2)
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