How do you find the derivative of the function #y = sin(cos(sinx))#?

3 Answers
Jun 25, 2018

Use the chain rule repeatedly

Explanation:

Use the chain rule.

Chain rule 1:

#d/dx[sincossinx]=coscossinx*d/dx[cossinx]#

Chain rule 2:

#d/dx[cossinx]=-sinsinx*d/dx[sinx]=-sinsinx*cosx#

Combine:

#d/dx[sincossinx]=coscossinx*(-sinsinx)*cosx#

So

#dy/dx=-cosxsinsinxcoscossinx#

Jun 25, 2018

#y'=-cos(cos(sin(x))*sin(sin(x))*cos(x)#

Explanation:

We must use the chain rule,
so we get

#cos(cos(sin(x)))*(-sin(sin(x))*cos(x)#

Jun 25, 2018

#d/dxsin(cos(sinx))=-cos(cos(sinx))sin(sinx)cosx#

Explanation:

To find derivatives of composite functions, we multiply the derivatives of each function with respect to its arguments.

So

#d/dxsin(cos(sinx))=(dsin(cos(sinx)))/(dcos(sinx)) *(d(cos(sinx)))/(dsinx) * (dsinx)/dx#

#=cos(cos(sinx))(-sin(sinx))cosx#