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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−(Π/8))

$$\mathrm{by}\:\mathrm{use}\:\mathrm{the}\:\mathrm{first}\: \\ $$$$\mathrm{principle},\mathrm{find} \\ $$$$\mathrm{dy}/\mathrm{dx}\:\mathrm{of}\: \\ $$$$\mathrm{y}=\mathrm{cos}\left(\mathrm{x}−\frac{\Pi}{\mathrm{8}}\right) \\ $$

Answered by icyfalcon999 last updated on 18/Aug/17

(dy/dx)=lim_(h→0) ((f(x+h)−f(x))/h) =lim_(h→0) ((cos (x+h−(π/8))−cos (x−(π/8)))/h) =lim_(h→0) ((cos (h+(x−(π/8)))−cos (x−(π/8)))/h) =lim_(h→0) ((cos (h)cos (x−(π/8))−sin h sin(x−(π/8)) −cos (x−(π/8)))/h) =lim_(h→0) ((cos (h)cos (x−(π/8))−cos (x−(π/8))−sin h sin(x−(π/8)) )/h) =lim_(h→0) ((cos (x−(π/8))(cos h−1)−sin h sin(x−(π/8)) )/h) =lim_(h→0) (((cos (x−(π/8))(cos h−1) )/h)−((sin h sin (x−(π/8)))/h)) =lim_(h→0) (((cos h−1 )/h)•cos (x−(π/8)))−lim_(h→0) (((sin h )/h)•sin (x−(π/8))) =lim_(h→0) ((cos h−1)/h)•lim_(h→0) cos (x−(π/8))−lim_(h→0) (((sin h )/h))•lim_(h→0) sin (x−(π/8)) =0•cos (x−(π/8))−1(sin (x−(π/8))) =−sin (x−(π/8))

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)−\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}} \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{h}−\frac{\pi}{\mathrm{8}}\right)−\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{h}} \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{h}+\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\right)−\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{h}} \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{h}\right)\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)−\mathrm{sin}\:\mathrm{h}\:\mathrm{sin}\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\:−\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{h}} \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{h}\right)\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)−\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)−\mathrm{sin}\:\mathrm{h}\:\mathrm{sin}\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\:}{\mathrm{h}} \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{cos}\:\mathrm{h}−\mathrm{1}\right)−\mathrm{sin}\:\mathrm{h}\:\mathrm{sin}\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\:}{\mathrm{h}} \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{cos}\:\mathrm{h}−\mathrm{1}\right)\:}{\mathrm{h}}−\frac{\mathrm{sin}\:\mathrm{h}\:\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{h}}\right) \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:\mathrm{h}−\mathrm{1}\:}{\mathrm{h}}\bullet\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\right)−\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{h}\:}{\mathrm{h}}\bullet\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\mathrm{h}−\mathrm{1}}{\mathrm{h}}\bullet\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)−\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{h}\:}{\mathrm{h}}\right)\bullet\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right) \\ $$$$=\mathrm{0}\bullet\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)−\mathrm{1}\left(\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$=−\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{8}}\right) \\ $$

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