show-that-f-x-sin-X-is-derivable-at-every-a-R-

Question Number 36703 by saikiran last updated on 04/Jun/18

s h o w t h a t f (x) = \sin X i s d e r i v a b l e a t e v e r y a ε R

Commented by abdo mathsup 649 cc last updated on 04/Jun/18

w e h a v e f o r a l l x \in R

{l i m}_{h \to 0} \frac{f (x + h) - f (x)}{h} = {l i m}_{h \to 0} \frac{s i n (x + h) - s i n x}{h}

= {l i m}_{h \to 0} \frac{s i n x c o s h + c o s x s i n h - s i n x}{h}

= {l i m}_{h \to 0} \frac{- s i n x (1 - c o s h)}{h} + \frac{s i n h}{h} c o s x b u t

w e k n o w t h a t {l i m}_{h \to 0} \frac{1 - c o s h}{h^{2}} = \frac{1}{2} \Rightarrow

{l i m}_{h \to 0} \frac{1 - c o s h}{h} = 0 a l s o {l i m}_{h \to 0} \frac{s i n h}{h} = 1 s o

{l i m}_{h \to 0} \frac{f (x + h) - f (x)}{h} = c o s x (\in R) s o f i s

d e r i v a b l a b e a t e v e r y p o i n t o f R .

Commented by MJS last updated on 04/Jun/18

to show

\lim_{h \to 0} \frac{\sin (x + h) - \sin (x - h)}{2 h} = \cos x

we can use \sin == \frac{e^{i x} - e^{- i x}}{2 i}; \cos x = \frac{e^{i x} + e^{- i x}}{2}

\lim_{h \to 0} \frac{e^{i (x + h)} - e^{- i (x + h)} - e^{i (x - h)} + e^{- i (x - h)}}{4 h i} =

= \lim_{h \to 0} \frac{\frac{d}{d h} (e^{i (x + h)} - e^{- i (x + h)} - e^{i (x - h)} + e^{- i (x - h)})}{\frac{d}{d h} (4 h i)} =

= \lim_{h \to 0} \frac{i}{4 i} (e^{i (x + h)} + e^{- i (x + h)} + e^{i (x - h)} + e^{- i (x - h)}) =

= \frac{1}{4} ({2 e}^{i x} + {2 e}^{- i x}) = \frac{e^{i x} + e^{- i x}}{2} = \cos x

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

1) m e a n i n g o f d e r i v a t i v e i s t h e a b i l i t y t o d r a w

t a n g e n t a t a p o i n t o n a c u r v e .

2) s i n x i s a c o n t i n u o u s c u r v e a t e v e r y p o i n t

a n d t a n g e n t c a n b e d r a w n a t e v e r y o i n t . s o i t

i s d e r i v a b l e

Answered by MJS last updated on 04/Jun/18

f (x) = \sin x

f^{'} (x) = \cos x

tangent in P = (\begin{matrix} p \\ \sin p \end{matrix}) :

y = x \cos p + \sin p - p \cos p

exists for all p \in R

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