How do you find the derivative of y= sin(sin(sin(x))) ?

Apr 23, 2018

$y ' = \cos x \cos \left(\sin x\right) \cos \left(\sin \left(\sin x\right)\right)$

Explanation:

Using the Chain Rule, we differentiate layer by player, first with the outermost sine.

We'll temporarily say

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

Then,

$y = \sin u$

$y ' = \cos u \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

To determine $\frac{\mathrm{du}}{\mathrm{dx}}$, look at $u = \sin \left(\sin x\right)$ and let $v = \sin x :$

$u = \sin v$

$\frac{\mathrm{du}}{\mathrm{dx}} = \cos v \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$

Well, $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \sin x = \cos x ,$ so

$\frac{\mathrm{du}}{\mathrm{dx}} = \cos v \cos x = \cos x \cos \left(\sin x\right)$

Thus,

$y ' = \cos u \cos x \cos \left(\sin x\right)$

$y ' = \cos x \cos \left(\sin x\right) \cos \left(\sin \left(\sin x\right)\right)$