Differentiation of cos cube x?

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Answer 1 · 2018-06-18T05:43:22

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$d/dx f( g(x) ) = g'(x) * f' (g(x) )$

In this case $d/dx cos^3 x = d/dx (cosx)^3$

$=> f(x) = x^3$ and $g(x) = cosx$

Use $d/dx cosx = -sinx$

$=> d/dx cos^3 x = 3 * cos^2 x * (-sinx )$

$=color(red)( -3sinxcos^2 x$