How do you differentiate $f (x) = \frac{{cos}^{4} x - x}{sec x}$ $f(x) = (cos^4 x - x) / sec x$ ?

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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,

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

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

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

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

How do you differentiate f(x) = (cos^4 x - x) / sec xf(x)=cos4x−xsecxf(x) = (cos^4 x - x) / sec x?