How do you find the derivative of y=sin^2 x?

Question

y = sin^(2) x
y equals sine squared x

How do you find the derivative of this? Please show process.

Answer 1

Chain rule

d/dx sin² x = 2 · sin x · cos x

Chain rule states:
d/dx f( g(x) ) = f'( g(x) ) · g'(x)


In the case of sin² x
f(x) = x²
g(x) = sin x

So, f( g(x) ) = sin² x


d/dx sin² x = d/dx f( g(x) ) = f'( g(x) ) · g'(x)

d/dx f(x) = d/dx x² = 2x
d/dx g(x) = d/dx sin x = cos x

d/dx sin² x = 2 sin x cos x

And through trigonometric identities, our solution can be changed further, if you so chose to use it.

sin (2x)

Answer 2

y = sinx * sinx
y = f(x)*g(x)
where it just happens that f(x) = g(x) = Sinx

chain rule for products:

y' = f'(x)g(x) + f(x)g'(x)

f(x) = sinx then f'(x) = cosx
(same for g(x), of course)

therefore:
y' = cosx sinx + sinx cosx
y' = 2 sinx cos x
y' = sin(2x)

or dy = sin(2x) dx

Answer 3

sin^2 x

= (sin x) (sin x)

So, ((sin x) (sin x)) '

= (sin x)' (sin x) + (sin x)' (sin x)

= (cos x) (sin x) + (cos x) (sin x)

= 2 cos x sin x

Answer 4

yeah that guy is right