what is the derivative of y= tan((sinx)^2)?

Answer 1

y = tan ( sin ² x )
let u = sin ² x
du/dx = 2 sin x cosx = sin 2x
y = tan u
dy/du = sec ² u = sec ² (sin ² x)
dy/dx = [ sec² (sin ² x) ] (sin 2x)

Answer 2

y' = sec^2 ((sin x)^2) * 2sin x* cos x
y' = sec^2 ((sin x)^2) * sin 2x

Answer 3

f(x) = tan( sin(x)^2 )
~~~~
u = sin(x)^2
du = 2sin(x)cos(x) dx
~~~~
f(x) = tan( u )
f(x)/du = 1/cos(u)^2
f(x)/du = 1/cos( sin(x)^2 )^2
~
f(x)/du * du/dx = f(x)/dx
~
f'(x) = 2sin(x)cos(x)/cos( sin(x)^2 )^2