Buktikan bahwa y= cos x , turunannya y= -sin x ?.... Question from @resnaadtya

October 2018 1 74 Report

Buktikan bahwa y= cos x , turunannya y= -sin x ?

Takamori37 Menggunakan metode limit.
$$\begin{align}y&=\cos x \\ y'&=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h} \\ &=\lim_{x\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h} \\ &=\lim_{h\to0}\frac{-\sin x\sin h}{h}+\lim_{h\to0}\frac{\cos x\cos h-\cos x}{h} \\ &=(-\sin x)\times\lim_{x\to0}\frac{\sin h}{h}+\cos x\times \lim_{x\to0}\frac{\cos h-1}{h} \\ &=(-\sin x)\times 1+\cos x\times 0 \\ &=-\sin x+0 \\ &=-\sin x \end{align}$

Untuk pembuktian: $\displaystyle \lim_{h\to0}\frac{\cos h-1}{h}=0$
$$\begin{align}\lim_{h\to0}\frac{\cos h-1}{h}&=\lim_{h\to0}\frac{\cos h-1}{h}\times\frac{\cos h+1}{\cos h+1} \\ &=\lim_{h\to0}\frac{\cos^2h-1}{h(\cos h+1)} \\ &=\lim_{h\to0}\frac{-(1-\cos^2h)}{h(\cos h+1)} \\ &=\lim_{h\to0}\frac{-\sin^2h}{h(\cos h+1)} \\ &=\lim_{h\to0}\frac{\sin h}{h}\times\lim_{h\to0}\frac{-\sin h}{\cos h+1} \\ &=1\times\frac{-\sin 0}{\cos 0+1} \\ &=\frac{-0}{1+1} \\ &=0 \end{align}$

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