by-use-the-first-principle-find-dy-dx-of-y-cos-x-8-

Question Number 19903 by j.masanja06@gmail.com last updated on 17/Aug/17

by use the first

principle, find

dy / dx of

y = \cos (x - \frac{Π}{8})

Answered by icyfalcon999 last updated on 18/Aug/17

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

= \lim_{h \to 0} \frac{\cos (x + h - \frac{π}{8}) - \cos (x - \frac{π}{8})}{h}

= \lim_{h \to 0} \frac{\cos (h + (x - \frac{π}{8})) - \cos (x - \frac{π}{8})}{h}

= \lim_{h \to 0} \frac{\cos (h) \cos (x - \frac{π}{8}) - \sin h \sin (x - \frac{π}{8}) - \cos (x - \frac{π}{8})}{h}

= \lim_{h \to 0} \frac{\cos (h) \cos (x - \frac{π}{8}) - \cos (x - \frac{π}{8}) - \sin h \sin (x - \frac{π}{8})}{h}

= \lim_{h \to 0} \frac{\cos (x - \frac{π}{8}) (\cos h - 1) - \sin h \sin (x - \frac{π}{8})}{h}

= \lim_{h \to 0} (\frac{\cos (x - \frac{π}{8}) (\cos h - 1)}{h} - \frac{\sin h \sin (x - \frac{π}{8})}{h})

= \lim_{h \to 0} (\frac{\cos h - 1}{h} ∙ \cos (x - \frac{π}{8})) - \lim_{h \to 0} (\frac{\sin h}{h} ∙ \sin (x - \frac{π}{8}))

= \lim_{h \to 0} \frac{\cos h - 1}{h} ∙ \underset{h \to 0}{\lim c o s} (x - \frac{π}{8}) - \lim_{h \to 0} (\frac{\sin h}{h}) ∙ \underset{h \to 0}{\lim s i n} (x - \frac{π}{8})

= 0 ∙ \cos (x - \frac{π}{8}) - 1 (\sin (x - \frac{π}{8}))

= - \sin (x - \frac{π}{8})