MCQ
Derivative of $\cos ^{-1}\left(4 x^3-3 x\right)$ :
  • A
    $\frac{3}{\sqrt{1-x^2}}$
  • $\frac{-3}{\sqrt{1-x^2}}$
  • C
    $\frac{1}{\sqrt{1+x^2}}$
  • D
    $-\frac{3}{\sqrt{1+x^2}}$

Answer

Correct option: B.
$\frac{-3}{\sqrt{1-x^2}}$
(B)Put $x=\cos \theta$$
\begin{array}{l}
\cos ^{-1}\left(4 x^3-3 x\right)=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right) \\
\Rightarrow \quad \cos ^{-1}(\cos 3 \theta)=3 \theta=3 \cos ^{-1} x[\because x=\cos \theta] \\
\therefore \quad \frac{d}{d x}\left(3 \cos ^{-1} x\right)=3 \frac{d}{d x}\left(\cos ^{-1} x\right)=3\left\{\frac{-1}{\sqrt{1-x^2}}\right\} \\
=\frac{-3}{\sqrt{1-x^2}}
\end{array}
$
Hence correct option is (B).

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