Question

How do you differentiate `sin^2(x/2)`?

Answer 1

By the chain rule.

Letting `y = sin^2(u)` and `u = x/2`, we need to differentiate both functions and multiply the derivatives together.

The derivative of `y = sin^2u` can be obtained as follows:

`y = (sinu)(sinu)`

By the product rule:

`y' = cosu xx sinu + cosu xx sinu`

`y' = 2cosusinu`

`y' = sin2u`

The derivative of `u = x/2` can be obtained using the quotient rule:

`u = x/2`

`u' = (1 xx 2 - x xx 0)/2^2`

`u' = 2/4`

`u' = 1/2`

So, the derivative of `y = sin^2(x/2)` is:

`dy/dx = sin2u xx 1/2`

`dy/dx = sin2(x/2) xx 1/2`

`dy/dx = 1/2 xx 2 xx sin(x/2) xx cos(x/2)`

`dy/dx = sin(x/2) xx cos(x/2)`

Hopefully this helps!

Answer 2

`1/2sinx`.

Explanation:

We use the Trigo. Identity `: 1-costheta=2sin^2(theta/2)`.

So, in our case, `sin^2(x/2)=1/2(1-cosx)`. Hence,

`d/dx[sin^2(x/2)]=d/dx[1/2(1-cosx)]=1/2d/dx[1-cosx]`

`=1/2[d/dx1-d/dxcosx]=1/2[0-(-sinx)]=1/2sinx`.

Note that, since, `sintheta=2sin(theta/2)cos(theta/2)`

`d/dx[sin^2(x/2)=1/2sinx=1/2(2sin(x/2)cos(x/2))=sin(x/2)cos(x/2),`

just to match With theAnswer furnished by, HSBC244 !

Enjoy Maths!