y = (sinx)x+sin−1√x
Let u = (sinx)x and v = sin−1√x
Now y = u + v

dy

dx

=

du

dx

+

dv

dx

... (i)
Consider u = (sinx)x
Taking logarithms on both the sides, we have,
logu = xlog (sin x)
Differentiating with respect to x, we have,

1

u

.

du

dx

= log (sin x) +

x

sinx

. cos x
⇒

du

dx

= (sinx)x (log (sin x) + x cot x) ... (ii)
Consider v = sin−1√x

dv

dx

=

1

√1−x

×

1

2√x

... (iii)
From (i), (ii) and (iii)
We get ,

dy

dx

= (sinx)x (log (sinx) + x cot x) +

1

2√x√1−x