How do you differentiate #f(x) = (cos^4 x - x) / sec x#?

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Answer 1 · 2017-08-04T11:23:43

# f'(x)=xsinx-5sinxcos^4x-cosx.#

We have, #f(x)=(cos^4x-x)/secx=(cos^4x-x)cosx,#

#:. f(x)=cos^5x-xcosx.#

#:. f'(x)=d/dx{(cosx)^5}-d/dx{xcosx}.#

Using the Chain Rule & Product Rule, we get,

#f'(x)=5(cosx)^(5-1)*d/dx{cosx}-[xd/dx{cosx}+cosx*d/dx{x}],#

#=5cos^4x*(-sinx)-[x(-sinx)+cosx*1],#

#=-5sinxcos^4x+xsinx-cosx,#

# rArr f'(x)=xsinx-5sinxcos^4x-cosx.#