How do you differentiate #f(x)=(cos x+ 4sin^2x^2)^6# using the chain rule?

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Answer 1 · 2018-06-02T13:15:14

#6(cosx+4sin^2x^2)^5(-sinx+16xsinx^2cosx^2)#

In several stages.

First stage - chain rule for the bracket as a whole:
#d/dx[(cosx+4sin^2x^2)^6]#
#=#
#6(cosx+4sin^2x^2)^5*d/dx[cosx+4sin^2x^2]#

Second stage - two terms, first one simple, second chain rule again:
#d/dx[cosx]=-sinx#
#d/dx[4sin^2x^2]=8sinx^2 *d/dx [sinx^2]#

Third stage - chain rule again
#d/dx[sinx^2]=cosx^2*d/dx[x^2]#

Fourth stage - simple differentiation
#d/dx[x^2]=2x#

Work back through the process:
#d/dx[x^2]=2x#
#d/dx[sinx^2]=2xcosx^2#
#d/dx[4sin^2x^2]=16xsinx^2cosx^2#

#d/dx[(cosx+4sin^2x^2)^6]#
#=#
#6(cosx+4sin^2x^2)^5(-sinx+16xsinx^2cosx^2)#