Question #d8431

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Answer 1 · 2017-02-17T13:03:06

$d/dx sec^3(sqrt(cosx)) = -(3sinxsec^3(sqrt(cosx)) tan (sqrt(cosx)) )/(2sqrt(cosx))$

$d/dx sec^3(sqrt(cosx)) = d/(du) u^3 * d/(dy) sec y * d/(dt) sqrt(t) d/dx cosx$

where:

$u= sec(sqrt(cosx))$

$y= sqrt(cosx)$

$t = cosx$

so:

$d/dx sec^3(sqrt(cosx)) = 3u^2 * secy tany * 1/(2sqrt(t)) * (-sinx)$

$d/dx sec^3(sqrt(cosx)) = 3sec^2(sqrt(cosx)) * sec(sqrt(cosx)) tan (sqrt(cosx)) * 1/(2sqrt(cosx)) * (-sinx)$

$d/dx sec^3(sqrt(cosx)) = -(3sinxsec^3(sqrt(cosx)) tan (sqrt(cosx)) )/(2sqrt(cosx))$