Differentiate the function y=sin^4(sqrt(u))

Here, you have one function inside another (inside
another) ; so you need to use the chain rule (twice).

Let x
= sqrt(u), f(x) = sin^4(x)

dy/du = df/dx *
dx/du

df/dx = 4sin^3(x) * cos(x)   --> here, you
used the chain rule to compute d/dx ( g^4), where g = sin x, and dg/dx = cos
x.

dx/du = d/du ( u^(1/2)) = 1/2
u^(-1/2)

So,

dy/du = 4sin^3(x)
* cos(x) * 1/2 u^(-1/2)

= 4sin^3(sqrt(u)) * cos(sqrt(u)) *
1/2 u^(-1/2)