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Pembuktian Turunan Cosec X = - Cotan X Cosec X

Darimana datangnya rumus turunan trigonometri Turunan Cosec X = - Cotan X.Cosec X, Atau Mengapa Turunan Cosec X = - Cotan X.Cosec X? Berikut pembuktian yang bisa dilakukan. (asumsi penurunan terhadap x).

Pembuktian ini akan menggunakan pendekatan limit, , $$ f^\prime (x) = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ \text {dengan catatan nilai limit harus ada} $$
Selain itu juga akan digunakan rumus trigonometri, $$ \sin (A+B) = \sin A \cos B + \cos A \sin B \\ \cot A = \frac{\cos A}{\sin A} \, $ dan $ \csc x A = \frac{1}{\sin A } $$
Perhatikan: $$ \text {misalkan} f(x) = \csc x \\ f(x) = \frac{1}{\sin x} \\ f(x+h) = \frac{1}{\sin (x+h)} \\ f(x+h) = \frac{1}{\sin x \cos h + \cos x \sin h} $$
Sekarang juga ingat rumus: $$ \cos 2A = 1 - 2\sin ^2 A \\ \text {jika 2A=h, maka} \\ \cos h = 1 - 2\sin ^2 \frac{1}{2} h \\ \cos h - 1 = - 2\sin ^2 \frac{1}{2} h \\ \cos h - 1 = - 2\sin ^2 \frac{1}{2} h \\ \cos h - 1= - 2\sin \frac{1}{2} h . \sin \frac{1}{2} h $$
Mari kita mulai mengunakan pendekatan limit,
$$ f^\prime (x) = \displaystyle \lim_{h \to 0 } \frac{f(x+h) - f(x) }{h} \\ = \displaystyle \lim_{h \to 0 } \frac{ \frac{1}{\sin x \cos h + \cos x \sin h} - \frac{1}{\sin x} }{h} \\ = \displaystyle \lim_{h \to 0 } \frac{ \frac{\sin x - (\sin x \cos h + \cos x \sin h) }{\sin x (\sin x \cos h + \cos x \sin h ) } }{h} \\ = \displaystyle \lim_{h \to 0 } \frac{ \frac{\sin x - \sin x \cos h - \cos x \sin h }{\sin x (\sin x \cos h + \cos x \sin h ) } }{h} $$
$$ = \displaystyle \lim_{h \to 0 } \frac{ \frac{\sin x (1 - \cos h ) - \cos x \sin h }{\sin x (\sin x \cos h + \cos x \sin h ) } }{h} \\ = \displaystyle \lim_{h \to 0 } \frac{ \frac{\sin x 2\sin \frac{1}{2} h . \sin \frac{1}{2} h - \cos x \sin h }{\sin x (\sin x \cos h + \cos x \sin h ) } }{h} \\ = \displaystyle \lim_{h \to 0 } \frac{ \frac{ \sin x 2\sin \frac{1}{2} h . \sin \frac{1}{2} h - \cos x \sin h }{h} }{\sin x (\sin x \cos h + \cos x \sin h ) } $$

$$ = \displaystyle \lim_{h \to 0 } \frac{ \frac{ \sin x 2\sin \frac{1}{2} h . \sin \frac{1}{2} h }{h} - \frac{ \cos x \sin h }{h} }{\sin x (\sin x \cos h + \cos x \sin h ) } \\ = \displaystyle \lim_{h \to 0 } \frac{ \sin x 2\sin \frac{1}{2} h . \frac{ \sin \frac{1}{2} h }{h} - \cos x \frac{ \sin h }{h} }{\sin x (\sin x \cos h + \cos x \sin h ) } \\ = \frac{ \displaystyle \lim_{h \to 0 } \sin x 2\sin \frac{1}{2} h . \displaystyle \lim_{h \to 0 } \frac{ \sin \frac{1}{2} h }{h} - \displaystyle \lim_{h \to 0 } \cos x \displaystyle \lim_{h \to 0 }\frac{ \sin h }{h} }{ \displaystyle \lim_{h \to 0 } \sin x (\sin x \cos h + \cos x \sin h ) } $$
$$ = \frac{ \sin x 2(\sin \frac{1}{2} . 0 ). \frac{1}{2} - \cos x . 1 }{ \sin x (\sin x \cos 0 + \cos x \sin 0 ) } \\ = \frac{ \sin x 2(\sin 0 ). \frac{1}{2} - \cos x }{ \sin x (\sin x . 1 + \cos x . 0 ) } \\ = \frac{ \sin x 2 . 0 . \frac{1}{2} - \cos x }{ \sin x (\sin x + 0 ) } \\ = \frac{ 0 - \cos x }{ \sin x (\sin x ) } \\ = \frac{ - \cos x }{ \sin x (\sin x ) } \\ = - \frac{ 1 }{ \sin x } . \frac{ \cos x }{ \sin x } \\ = - \csc x \cot x $$



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