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

Sep 18, 2016

By the chain rule.

Letting $y = {\sin}^{2} \left(u\right)$ and $u = \frac{x}{2}$, we need to differentiate both functions and multiply the derivatives together.

The derivative of $y = {\sin}^{2} u$ can be obtained as follows:

$y = \left(\sin u\right) \left(\sin u\right)$

By the product rule:

$y ' = \cos u \times \sin u + \cos u \times \sin u$

$y ' = 2 \cos u \sin u$

$y ' = \sin 2 u$

The derivative of $u = \frac{x}{2}$ can be obtained using the quotient rule:

$u = \frac{x}{2}$

$u ' = \frac{1 \times 2 - x \times 0}{2} ^ 2$

$u ' = \frac{2}{4}$

$u ' = \frac{1}{2}$

So, the derivative of $y = {\sin}^{2} \left(\frac{x}{2}\right)$ is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin 2 u \times \frac{1}{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin 2 \left(\frac{x}{2}\right) \times \frac{1}{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \times 2 \times \sin \left(\frac{x}{2}\right) \times \cos \left(\frac{x}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin \left(\frac{x}{2}\right) \times \cos \left(\frac{x}{2}\right)$

Hopefully this helps!

Sep 18, 2016

$\frac{1}{2} \sin x$.

#### Explanation:

We use the Trigo. Identity $: 1 - \cos \theta = 2 {\sin}^{2} \left(\frac{\theta}{2}\right)$.

So, in our case, ${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1}{2} \left(1 - \cos x\right)$. Hence,

$\frac{d}{\mathrm{dx}} \left[{\sin}^{2} \left(\frac{x}{2}\right)\right] = \frac{d}{\mathrm{dx}} \left[\frac{1}{2} \left(1 - \cos x\right)\right] = \frac{1}{2} \frac{d}{\mathrm{dx}} \left[1 - \cos x\right]$

$= \frac{1}{2} \left[\frac{d}{\mathrm{dx}} 1 - \frac{d}{\mathrm{dx}} \cos x\right] = \frac{1}{2} \left[0 - \left(- \sin x\right)\right] = \frac{1}{2} \sin x$.

Note that, since, $\sin \theta = 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)$

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!