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

1 Answer
Apr 23, 2018

y'=cosxcos(sinx)cos(sin(sinx))

Explanation:

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

We'll temporarily say

u=sin(sinx)

Then,

y=sinu

y'=cosu*(du)/dx

To determine (du)/dx, look at u=sin(sinx) and let v=sinx:

u=sinv

(du)/dx=cosv*(dv)/dx

Well, (dv)/dx=d/dxsinx=cosx, so

(du)/dx=cosvcosx=cosxcos(sinx)

Thus,

y'=cosucosxcos(sinx)

y'=cosxcos(sinx)cos(sin(sinx))