Question:
Mark $(\sqrt{ })$ against the correct answer in the following:
If $y=\cos ^{-1}\left(4 x^{3}-3 x\right)$ then $\frac{d y}{d x}=$ ?
A. $\frac{3}{\sqrt{1-x^{2}}}$
B. $\frac{-3}{\sqrt{1-\mathrm{x}^{2}}}$
C. $\frac{4}{\sqrt{1-x^{2}}}$
D. $\frac{4}{\left(3 x^{2}-1\right)}$
Solution:
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-3}{\sqrt{1-\mathrm{x}^{2}}}$