Given, y=(sin‌x)x+sin−1x . . . (i) Let ‌‌u=(sin‌x)x . . . (ii) Then, Eq. (i) becomes, y=u+sin−1√x. . . (iii) On taking log both sides of Eq. (ii), we get log‌u=x‌log‌sin‌x On differentiating both sides w.r.t., x, we get ‌
1
u
‌
du
dx
‌‌=x‌
d
dx
(log‌sin‌x)+log‌sin‌x‌
d
dx
(x) ⇒‌
du
dx
‌‌=u[x×‌
1
sin‌x
‌
d
dx
(sin‌x)+log‌sin‌x(1)] ⇒‌‌‌
du
dx
‌‌=(sin‌x)x[‌
x
sin‌x
×cos‌x+log‌sin‌x] ⇒‌
du
dx
‌‌=(sin‌x)x[x‌cot‌x+log‌sin‌x] . . . (iv) On differentiating both sides of Eq. (iii) w.r.t. x, we get ‌‌