[tex] \frac{d(cotan \: x)}{dx} = - {cosec}^{2} x[/tex]
tolong bantu pembuktiannya Kaka
Jawaban:
[tex] \frac{d(cotan \: x)}{dx} [/tex]
[tex] = \frac{d}{dx} ( \frac{cos \: x}{sin \: x}) [/tex]
[tex] = \frac{ \frac{d(cos \: x)}{dx} \times sin \: x - cos \: x \times \frac{d(sin \: x)}{dx} }{ {sin}^{2} \: x} [/tex]
[tex] = \frac{ - sin \: x \times sin \: x - cos \: x \times cos \: x}{ {sin}^{2} \: x } [/tex]
[tex] = \frac{ - ( {sin}^{2} \: x + {cos}^{2} \: x) }{ {sin}^{2} \: x} [/tex]
[tex] = - \frac{1}{ {sin}^{2} \: x } [/tex]
[tex] = - {cosec}^{2} \: x[/tex]
[tex]\displaystyle\frac{d}{dx}(cot(x))=\frac{d}{dx}\left(\frac{cos(x)}{sin(x)}\right)[/tex]
u = cos(x) v = sin(x)
[tex]\displaystyle\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}\\\\\frac{d}{dx}\left(\frac{cos(x)}{sin(x)}\right)=\frac{cos(x)'sin(x)-cos(x)sin(x)'}{sin^2(x)}\\\\=\frac{-sin(x)sin(x)-cos(x)cos(x)}{sin^2(x)}\\\\=\frac{-sin^2(x)-cos^2(x)}{sin^2(x)}=\frac{-(sin^2(x)+cos^2(x))}{sin^2(x)}\\\\=\frac{-1}{sin^2(x)}=-\frac{1}{sin^2(x)}=\bf -csc^2(x)~~\rm terbukti[/tex]
(xcvi)
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