Question

If $u={\sin }^{-1}(x-y),x=3t,y=4t^3,$ then $\frac{du}{dt}=$

• A. $\frac{3}{\sqrt{(1-t^2)}}$
• B. $\frac{3}{\sqrt{(1+t^2)}}$
• C. $\frac{2}{\sqrt{(1-t^2)}}$
• D. $\frac{4}{\sqrt{(1-t^2)}}$

Answer 1

The correct option is A $\frac{3}{\sqrt{(1-t^2)}}$

Given $u={\sin }^{-1}(x-y),x=3t,y=4t^3$

$\Rightarrow \sin u=x-y=3t-4t^3=3\sin \theta -4{\sin }^3\theta$

Put $t=\sin \theta$

$\Rightarrow \sin u=3\sin \theta -4{\sin }^3\theta =\sin 3\theta$

or $u=3\theta =3{\sin }^{-1}t$

$\therefore \frac{du}{dt}=\frac{3}{\sqrt{1-t^2}}$