How do you find the derivative of #y= sin{cos^2(tanx)}#?

1 Answer
Nov 8, 2017

Answer:

#(dy)/(dx)=-2sec^2(x)sin(tanx)cos(tanx)cos(cos^2(tan(x)))#

Explanation:

#y= sin(cos^2(tanx))=sin(f(x))#

#y=sin(f(x))=>(dy)/(dx)=f'(x)cos(f(x))#

#f(x)=cos^2(g(x))=(cos(g(x)))^2#
#f'(x)=2*d/(dx)[cos(g(x))] * cos(g(x))#
#f'(x)=2*-g'(x)sin(g(x))*cos(g(x))#
#f'(x)=-2g'(x)sin(g(x))cos(g(x))#

#g(x)=tanx#
#g'(x)=sec^2x#

Therefore, #f'(x)=-2sec^2(x)sin(tanx)cos(tanx)#

#(dy)/(dx)=-2sec^2(x)sin(tanx)cos(tanx)cos(cos^2(tan(x)))#