# How do you find the derivative of sin^2(8x)-(pi)x?

Aug 13, 2017

$\frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = 8 \sin \left(16 x\right) - \pi$

#### Explanation:

We have: ${\sin}^{2} \left(8 x\right) - \left(\pi\right) x$

$= {\left(\sin \left(8 x\right)\right)}^{2} - \pi x$

First, let's apply the difference rule:

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2}\right) - \frac{d}{\mathrm{dx}} \left(\pi x\right)$

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2}\right) - \pi$

Then, let's use the chain rule.

Let $u = \sin \left(8 x\right) R i g h t a r r o w u ' = \cos \left(8 x\right)$, $v = {u}^{2} R i g h t a r r o w v ' = 2 u$ and $w = 8 x R i g h t a r r o w w ' = 8$:

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = u ' \cdot v ' \cdot w ' - \pi$

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = \cos \left(8 x\right) \cdot 2 u \cdot 8 - \pi$

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = 8 \cdot 2 u \cos \left(8 x\right) - \pi$

Now, let's replace $u$ with $\sin \left(8 x\right)$:

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = 8 \cdot 2 \sin \left(8 x\right) \cos \left(8 x\right) - \pi$

Let's apply the double angle formula for $\sin \left(x\right)$; $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$:

$R i g h t a r r o w \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = 8 \cdot \sin \left(2 \cdot 8 x\right) - \pi$

$\therefore \frac{d}{\mathrm{dx}} \left({\left(\sin \left(8 x\right)\right)}^{2} - \pi x\right) = 8 \sin \left(16 x\right) - \pi$