What is the derivative of y=(sin^2)*x?

Question

Using the chain rule?

Answer 1

This is the way the chain rule works. For any function u

du/dx = u' du/dx.

So if u(x) = sin(x)

d(u²)/dx = u' du/dx = 2u du/dx

du/dx = dsin(x)/dx = cos(x)

So the result is 2sin(x)cos(x) = sin(2x)

Answer 2

i THINK (im in this same class doing the same thing)
that the answer issss...
2sinx*cosx
because the differentiation of sin^2 is 2sinx and
the differentiation of sinx is cosx

Answer 3

y = sin ² x
y = (sin x) ²
dy/dx = 2 ( sin x )( cos x )
dy/dx = sin 2x

Answer 4

if you mean y = (sinx)^2 then
let z = sinx
then y = z^2
(dy/dz)*(dz/dx) = 2z * cosx = 2 sinx cosx