Turunan Fungsi Aturan Pembagian


Sifat : Jika f dan g fungsi-fungsi yang terdiferensial maka \left ( \frac{f}{g} \right )(x) =  \frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} yakni Dx \frac{f(x)}{g(x)}= \frac{D_x[f(x)]g(x)-f(x)D_x[g(x)]}{g^2(x)}

Bukti :

Andaikan : F(x) = \frac{f(x)}{g(x)}

F'(x) = lim_{h \to 0} \frac{F(x+h)-F(x)}{h}

= lim_{h \to 0} \frac{\frac{f(x+h)}{g(x+h)}- \frac{f(x)}{g(x)}}{h}

= lim_{h \to 0} \frac{g(x)f(x+h)-f(x)g(x+h)}{h}. \frac{1}{g(x)g(x+h)}

= lim_{h \to 0} (\frac{g(x)f(x+h)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{h}. \frac{1}{g(x)g(x+h)})

= lim_{h \to 0} (\frac{g(x)(f(x+h)-f(x))+f(x)(g(x)-g(x+h))}{h}. \frac{1}{g(x)g(x+h)})

= lim_{h \to 0} ((g(x)\frac{f(x+h)-f(x)}{h}+f(x)\frac{g(x)-g(x+h)}{h}).\frac{1}{g(x)g(x+h)})

= lim_{h \to 0} ((g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h}).\frac{1}{g(x)g(x+h)})

= (lim_{h \to 0} g(x)\frac{f(x+h)-f(x)}{h} lim_{h \to 0} f(x)\frac{g(x+h)-g(x)}{h}) (lim_{h \to 0} \frac{1}{g(x)g(x+h)})

= (lim_{h \to 0} \quad g(x).lim_{h \to 0} \frac{f(x+h)-f(x)}{h} lim_{h \to 0} \quad f(x).lim_{h \to 0} \frac{g(x+h)-g(x)}{h}) (lim_{h \to 0} \frac{1}{g(x)g(x+h)})

= (f'(x)g(x)-f(x)g'(x)) (\frac{1}{g(x)g(x)})

= \frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} \blacksquare

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7 comments on “Turunan Fungsi Aturan Pembagian

  1. Ping-balik: Kuy Belajar Matematika !!!

  2. Ping-balik: Pos blog pertama – Kuy Belajar Matematika !!!

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