# What is the derivative of? : sin^2(x/2) \ cos^2(x/2)

Jun 22, 2017

$\left({\sin}^{2} \left(\frac{x}{2}\right) \cdot {\cos}^{2} \left(\frac{x}{2}\right)\right) ' = \left(- \cos \left(\frac{x}{2}\right) + 2 {\cos}^{3} \left(\frac{x}{2}\right)\right) \sin \left(\frac{x}{2}\right)$

#### Explanation:

Note that ${\sin}^{2} \left(\frac{x}{2}\right) = 1 - {\cos}^{2} \left(\frac{x}{2}\right)$

So ${\sin}^{2} \left(\frac{x}{2}\right) \cdot {\cos}^{2} \left(\frac{x}{2}\right) = \left(1 - {\cos}^{2} \left(\frac{x}{2}\right)\right) \cdot {\cos}^{2} \left(\frac{x}{2}\right)$
$= {\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{4} \left(\frac{x}{2}\right)$
Finding the derivative of this is a little simpler :)

Let's first find the derivative of ${\cos}^{2} \left(\frac{x}{2}\right)$ and use the same method for the other part.

To this use the chain rule.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
Let's say $u = \cos \left(\frac{x}{2}\right)$
So ${\cos}^{2} \left(\frac{x}{2}\right) = {u}^{2}$

$\frac{d}{\mathrm{du}} \left({u}^{2}\right) = 2 u$

However to calculate $\frac{\mathrm{du}}{\mathrm{dx}}$, we would have to use the chain rule again. I'm going to skip that, but just note that it's nescesary.

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

Bringing it all together:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 u \cdot \left(- \frac{1}{2} \sin \left(\frac{x}{2}\right)\right)$
$2 \cos \left(\frac{x}{2}\right) \cdot \left(- \frac{1}{2} \sin \left(\frac{x}{2}\right)\right) = - \cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)$

Using the same method for ${\cos}^{4} \left(\frac{x}{2}\right)$
$u = \cos \left(\frac{x}{2}\right)$
$\frac{d}{\mathrm{du}} {u}^{4} = 4 {u}^{3}$
$\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{2} \sin \left(\frac{x}{2}\right)$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\cos}^{3} \left(\frac{x}{2}\right) \cdot \left(- \frac{1}{2} \sin \left(\frac{x}{2}\right)\right)$
$= - 2 {\cos}^{3} \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)$

Now combine these in the final step.
$- \cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right) - \left(- 2 {\cos}^{3} \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)\right)$
$= \left(- \cos \left(\frac{x}{2}\right) + 2 {\cos}^{3} \left(\frac{x}{2}\right)\right) \sin \left(\frac{x}{2}\right)$

You could probably reduce this, but I think it's fine :)

Jun 23, 2017

$\frac{d}{\mathrm{dx}} {\sin}^{2} \left(\frac{x}{2}\right) \setminus {\cos}^{2} \left(\frac{x}{2}\right) = \frac{1}{4} \sin 2 x$

#### Explanation:

Let:

$y = {\sin}^{2} \left(\frac{x}{2}\right) \setminus {\cos}^{2} \left(\frac{x}{2}\right)$

We could apply the product rule and chain rule but the expression can be significantly simplified using the sine double angle formula:

$\sin 2 A = 2 \sin A \cos A \iff \sin A \cos A = \frac{1}{2} \sin 2 A$

Thus we can write the initial expression as:

$y = {\left(\frac{1}{2} \sin x\right)}^{2}$
$\setminus \setminus = \frac{1}{4} {\sin}^{2} x$

So differentiating, using the chain rule, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{4}\right) \left(2 \sin x\right) \left(\cos x\right)$
$\text{ } = \left(\frac{1}{4}\right) \left(2 \sin x \cos x\right)$

Again using the sine double angle formula, we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{4}\right) \sin \left(2 x\right)$