# What is the derivative of sin(x(pi/8))?

Jan 10, 2018

$y ' = \frac{d}{\mathrm{dx}} \left[\sin \left(\frac{\pi x}{8}\right)\right] = \frac{\pi}{8} \cos \left(\frac{\pi x}{8}\right)$

#### Explanation:

Apply the Chain Rule:

y'=d/dx[f(g(x)]=f'(g(x))*g'(x)

Given: $\sin \left(x \left(\frac{\pi}{8}\right)\right)$, We can rewrite this as $\sin \left(\frac{\pi x}{8}\right)$

Let $f \left(x\right) = \sin x$ and $g \left(x\right) = \frac{\pi x}{8}$

Thus, $f ' \left(x\right) = \cos x$ and $g ' \left(x\right) = \frac{\pi}{8}$

So,

$y ' = \frac{d}{\mathrm{dx}} \left[\sin \left(\frac{\pi x}{8}\right)\right] = \cos \left(\frac{\pi x}{8}\right) \cdot \frac{\pi}{8}$