Answer
(a) $dy$ = $-\frac{1}{(1+3u)^{2}}du$
(b) $dy$ = $2θ^{2}cos(2θ)+2θsin(2θ)$
Work Step by Step
(a)
$\frac{dy}{du}$ = $\frac{(1+3u)(2)-(1+2u)(3)}{(1+3u)^{2}}$
$\frac{dy}{du}$ = $\frac{2+6u-3-6u}{(1+3u)^{2}}$
$dy$ = $-\frac{1}{(1+3u)^{2}}du$
(b)
$\frac{dy}{dθ}$ = $2θ^{2}cos(2θ)+sin(2θ)2θ$
$dy$ = $2θ^{2}cos(2θ)+2θsin(2θ)$
