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Wednesday, January 10, 2024

Problem Set 2.5 - Chain Rule

 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 })

Sunday, January 7, 2024

Problem Set 2.4 Derivatif Trigonometri

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 

--------

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}




Problem Set 1.4 : Limit Involving Trigonometric Functions

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}