Problem Set 2.4 Derivatif Trigonometri

catatan :

$\frac{d}{dx} sin x~= cos\, x$

$D_x (cos x)~= - sin\, x$

$\frac{d}{dx}tanx~=sec^2 x$

$D_x (cot\, x)~=-csc\, x$

$\frac{d}{dx} sec x~=sec\, x\, tan\, x$ 

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$6. ~y = csc\, x$

$D_x(csc \, x) = D_x \frac{1}{sin x}$

$= \frac{sin x\, D_x(1) - (1)\, D_x(sin x)}{sin^2 x}$

$= \frac{0 - cos\, x}{sin^2 x}~=\frac{-1}{sin x}.\frac{cos\, x}{sin\, x}~= -csc\,x \,cot\, x$

$10.~y= \frac{sin x + cos x}{tan x}$ 

$\frac{d}{dx}(\frac{sin x + cos x}{tan x})$

$=\frac{tanx\, D_x (sin x+ cos x) - (sin x+ cos x)\, D_x (tan x)}{tan^2 x}$

 $= \frac{tanx(cosx-sinx)-(sinx+cosx)sec^2 x}{tan^2x}$

$= \frac{tanx(cosx-sinx)-(sinx+cosx)sec^2 x}{\frac{sin^2x}{cos^2x}}$ 

$= [sin x - \frac{sin^2 x}{cos x} - \frac{sin x}{cos^2 x} - \frac{1}{cos x}]  : [\frac{sin^2 x}{cos^2 x}]$

$= [sin x - \frac{sin^2 x}{cos x} - \frac{sin x}{cos^2 x} - \frac{1}{cos x}]  [\frac{cos^2 x}{sin^2 x}]$

$=\frac{cos^2 x}{sin x} - cos x - \frac{1}{sin x} - \frac{cos x}{sin^2 x}$