16. $ y= cos^3(\frac{x^2}{1 - x}) $
$ v= \frac{x^2}{1-x}, ~ u=cos \, v, ~ y = u^3 $
$\frac{dy}{dx} = \frac{dv}{dx}. \frac{du}{dv}. \frac{dy}{du}$
$ = (\frac{(1 - x)D_x(x^2) - (x^2) D_x(1 - x)}{(1 - x)^2}). (- sin \, v).(3u^2) $
$ = \frac{(1-x)(2x) -(x^2)(-1)}{(1 - x)^2}. sin(\frac{x^2}{1 -x}). (-3 cos^2 (\frac{x^2}{1 - x}))$
$ = \frac{-3(2x - x^2)}{(1 - x)^2}cos^2(\frac{x^2}{1-x}) sin(\frac{x^2}{1 -x }) $
catatan :
$ \frac{d}{dx} sin x~= cos\, x$
$ D_x (cos x)~= - sin\, x$
$ \frac{d}{dx}tanx~=sec^2 x$
$D_x (cot\, x)~=-csc\, x $
$\frac{d}{dx} sec x~=sec\, x\, tan\, x$
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$ 6. ~y = csc\, x$
$ D_x(csc \, x) = D_x \frac{1}{sin x}$
$ = \frac{sin x\, D_x(1) - (1)\, D_x(sin x)}{sin^2 x}$
$ = \frac{0 - cos\, x}{sin^2 x}~=\frac{-1}{sin x}.\frac{cos\, x}{sin\, x}~= -csc\,x \,cot\, x$
$10.~y= \frac{sin x + cos x}{tan x} $
$\frac{d}{dx}(\frac{sin x + cos x}{tan x})$
$=\frac{tanx\, D_x (sin x+ cos x) - (sin x+ cos x)\, D_x (tan x)}{tan^2 x}$
$= \frac{tanx(cosx-sinx)-(sinx+cosx)sec^2 x}{tan^2x}$
$= \frac{tanx(cosx-sinx)-(sinx+cosx)sec^2 x}{\frac{sin^2x}{cos^2x}}$
$ = [sin x - \frac{sin^2 x}{cos x} - \frac{sin x}{cos^2 x} - \frac{1}{cos x}] : [\frac{sin^2 x}{cos^2 x}]$
$ = [sin x - \frac{sin^2 x}{cos x} - \frac{sin x}{cos^2 x} - \frac{1}{cos x}] [\frac{cos^2 x}{sin^2 x}]$
$ =\frac{cos^2 x}{sin x} - cos x - \frac{1}{sin x} - \frac{cos x}{sin^2 x} $
$ 7. \lim_{\theta \to 0} \frac{sin\, 3\theta}{tan \, \theta} $
$= \lim_{\theta \to 0} \frac{sin \, 3\theta}{\frac{sin \theta}{cos \theta}} $
$ = \lim_{\theta \to 0} \frac{cos\theta \, sin 3\theta}{sin \theta} $
$ = \lim_{\theta \to 0} [ \frac{sin 3\theta}{3 \theta}. 3\, cos \theta . \frac{1}{sin \theta} . \theta ] $
$ = 3 \lim_{\theta \to 0} [ \frac{sin 3\theta}{3\theta}. cos \theta. \frac{\theta}{sin \theta}] $
$ = 3 (1.1.1)~=~3 $
$ 8. \lim_{\theta \to 0} \frac{tan 5\theta}{sin 2\theta} $
$ = \lim_{\theta \to 0} \frac{\frac{sin 5\theta}{cos 5\theta}}{sin 2\theta} $
$ = lim_{\theta \to 0} \frac{sin5 \theta}{ cos 5\theta \, sin 2\theta} $
$ = lim_{\theta \to 0} [\frac{sin 5\theta}{5\theta}.\frac{2\theta}{sin 2\theta}.\frac{1}{cos 5\theta}.5\theta. \frac{1}{2\theta}] $
$ = lim_{\theta \to 0} [\frac{sin 5\theta}{5\theta}.\frac{2\theta}{sin 2\theta}.\frac{1}{cos 5\theta}.\frac{5}{2}] $
$ = \frac{5}{2} lim_{\theta \to 0} [\frac{sin 5\theta}{5\theta}.\frac{2\theta}{sin 2\theta}.\frac{1}{cos 5\theta}] $
$ = \frac{5}{2} (1.1.1)~=~ \frac{5}{2} $
$9. \lim_{\theta \to 0} \frac{cot\, \pi \theta \, sin \theta}{2 sec \theta}$
$ = \lim_{\theta \to 0} \frac{\frac{cos \pi \theta}{sin \pi \theta} sin \theta}{\frac{2}{cos \theta}}$
$ = \lim_{\theta \to 0} \frac{cos \pi \theta \, sin \theta \, cos \theta}{2 sin \pi \theta} $
$= \lim_{\theta \to 0} [\frac{sin \theta}{\theta}. \frac{\pi \theta}{sin \pi \theta}. \frac{1}{\pi}. \frac{cos \pi \theta \, cos \theta}{2}]$
$ = \frac{1}{2\pi} ~\lim_{\theta \to 0} [\frac{sin \theta}{\theta}. \frac{\pi \theta}{sin \pi \theta}. cos \pi \theta \, cos \theta ]$
$ =\frac{1}{2 \pi}(1.1.1) ~=~\frac{1}{2 \pi}$