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

Question

16833 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-23T01:45:32

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

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))#