(a) Differentiate y = sin^-1(3 x -4 x^3) w.r.t. x, if x ∈[-1/2, 1/2].

$\displaystyle \mathrm{x}=\operatorname{sint}$ gives $\displaystyle \mathrm{y}=\sin ^{-1}(\sin 3 t)=3 t=3 \sin ^{-1} \mathrm{x}$

\[\frac{d \mathrm{y}}{d \mathrm{x}}=\frac{3}{\sqrt{1-\mathrm{x}^{2}}} \]

Aliter: $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3-12 \mathrm{x}^{2}}{\sqrt{1-\left(3 \mathrm{x}-4 \mathrm{x}^{3}\right)^{2}}}$

$\displaystyle \mathrm{x}=\operatorname{tant}$ gives $\displaystyle \mathrm{y}=\cos ^{-1}(\cos 2 t)=2 t=2 \tan ^{-1} \mathrm{x}$

\[\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{1+\mathrm{x}^{2}} \]

Aliter: $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sqrt{1-\left(\dfrac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)^{2}}} \cdot \frac{-4 \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}}$.