# Thread: finding the derivative of sin^2(x)

1. ## finding the derivative of sin^2(x)

I need help finding the derivative of:

sin^2(x)

I believe I have to use the half angle identity...

if this is true, then would it be:
1/2 (1-cos(2x)) ?

would I need to take the derivative of this after the identity?
If this holds true, would be 2sin(2x) ???

2. You can use the product rule straight-away: $\dfrac{d}{dx}\left(\sin^2{x}\right) = \dfrac{d}{dx}\left(\sin{x}\sin{x}\right) = \dfrac{d}{dx}\left(\sin{x}\right)\sin{x}+\dfrac{d} {dx}\left(\sin{x}\right)\sin{x}$ $= \cos{x}\sin{x}+\cos{x}\sin{x} = 2\sin{x}\cos{x} = 2\sin{2x}.$

3. Originally Posted by RET80
I need help finding the derivative of:

sin^2(x)

I believe I have to use the half angle identity...

if this is true, then would it be:
1/2 (1-cos(2x)) ?

would I need to take the derivative of this after the identity?
If this holds true, would be 2sin(2x) ???
1/2(1- cos(2x))= 1/2- 1/2 cos(2x). Of course, the derivative of 1/2 is 0. Using the chain rule, the derivative of cos(2x) is - sin(2x)(2x)'= -2 sin(2x). Of course, then, the derivative of -1/2 cos(2x) is (-1/2)(-2) sin(2x)= sin(2x). The derivative of 1/2(1- cos(2x)) is sin(2x), not 2 sin(2x).

But I would have done this using the chain rule directly- Let u= sin(x) and y= u^2= sin^2(x). Then $\frac{dy}{dx}= \frac{du^2}{dy}\frac{du}{dx}$ $= 2u (cos(x)= 2sin(x)cos(x)$.

Or, more simply, since the derivative of $x^2$ is 2x, the derivative of $sin^2(x)$ is 2x, the derivative of $sin^2(x)$ is 2 sin(x) times the derivative of sin(x), which is cos(x).

Of course, those are the same answer- sin(2x)= 2 sin(x)cos(x).