49. If \( x=\sin ^{-1}\left(3 t-4 t^{3}\right) \) and \( y=\cos ^{-1}\left(\sqrt{1-t^{2}}\right) \), then \( \frac{d y}{d x} \) is equal to (a) \( \frac{1}{2} \) (b) \( \frac{2}{5} \) (c) \( \frac{3}{2} \) (d) \( \frac{1}{3} \)

Question

 If \( x=\sin ^{-1}\left(3 t-4 t^{3}\right) \) and \( y=\cos ^{-1}\left(\sqrt{1-t^{2}}\right) \), then \( \frac{d y}{d x} \) is equal to

 (a) \( \frac{1}{2} \) (b) \( \frac{2}{5} \) (c) \( \frac{3}{2} \) (d) \( \frac{1}{3} \)

Answer 1

Option (d)i is correct

Let  t = sinx

Then x= arcsin(3sinz-4sin³z)

=,>x=arcsinsin3z=3z

So. dx/dz =3

Again y =arccos cosz=z

dy/d z =1

Hence dy/dx=1/3