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Answer 1 · 2017-10-12T03:43:02

Substitution and chain rule

Let $theta =3x^2$ and $p = cos theta$ then

$y = sin^3(cos(3x^2))$
$y = sin^3p$

For $y'$ we can use chain rule as

$(dy)/(dx) = (dy)/(dp)(dp)/(d theta)(d theta)/(dx)$

$(dy)/(dp) = 3sin^2pcosp$

$(dp)/(d theta) = -sin theta$

$(d theta)/dx = 6x$

$(dy)/(dx) = 3sin^2pcosp\times(-sin theta)\times6x$

$= -18xsin^2(cos(3x^2))cos(cos(3x^2))\times sin(3x^2)$
$=-9sin(2cos(3x^2))sin(cos(3x^2)) sin(3x^2)$