Find Derivative of y=sin(cos(sinx))

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Homework Help Overview

The discussion revolves around finding the derivative of the function y=sin(cos(sinx)), focusing on the application of differentiation rules, particularly the chain rule.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of the chain rule, suggesting the introduction of intermediate variables to simplify differentiation. There are questions regarding the correctness of the initial attempts and whether all necessary factors have been accounted for in the derivative.

Discussion Status

There is an ongoing exploration of the derivative with various participants providing feedback on each other's attempts. Some express uncertainty about the correctness of their approaches, while others affirm the validity of the chain rule application.

Contextual Notes

Participants are navigating through potential misunderstandings of differentiation rules and the complexity of the function involved. There is an acknowledgment of the possibility of applying rules incorrectly.

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Homework Statement


find the derivative using the appropriate rules

y=sin(cos(sinx))

The Attempt at a Solution



d/dx[sin(cos(sinx))] = [cos(cos(sinx))] * [-sin(sinx)] =

-cos(cos(sinx))(sin(sinx))

Is that right?

It doesn't feel right.
 
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Try making an extended chain rule...let t=sinx and then u=cost ...and then differentiate and use the chain rule
 
you're very close! 1 more angle
 
is the first factor done? and -sin(sinx) needs more?
 
something like this: y= sinu ; u=cos(t) ; t=sinx

d/du[y] * d/dt * d/dx[t] ?
 
Last edited:
= cosu * (-sint) * cosx =

-sin(sinx)*cos(cos(sinx))*cosx ??
 
well the chain rule is correct...so I would assume that it is correct
 
It wouldn't be uncommon for me to apply a rule incorrectly.
 
Well it is correct
 

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