Question-143134

Question Number 143134 by help last updated on 10/Jun/21

Answered by liberty last updated on 10/Jun/21

y′=lim_(h→0) ((cos ((x+h)^2 −4(x+h))−cos (x^2 −4x))/h) =lim_(h→0) ((cos (x^2 +2hx+h^2 −4x−4h)−cos (x^2 −4x))/h) = lim_(h→0) ((−2sin (((2x^2 −8x+2hx+h^2 −4h)/2))sin (((2hx+h^2 −4h)/2)))/h) = −2sin (x^2 −4x).lim_(h→0) ((sin h(((2x+h−4)/2)))/h) =−2sin (x^2 −4x)(((2x−4)/2)) =−2(x−2)sin (x^2 −4x) =(4−2x)sin (x^2 −4x)

$$\:{y}'=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\left({x}+{h}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{h}\right)\right)−\mathrm{cos}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left({x}^{\mathrm{2}} +\mathrm{2}{hx}+{h}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}{h}\right)−\mathrm{cos}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)}{{h}} \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:\left(\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{2}{hx}+{h}^{\mathrm{2}} −\mathrm{4}{h}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}{hx}+{h}^{\mathrm{2}} −\mathrm{4}{h}}{\mathrm{2}}\right)}{{h}} \\ $$$$=\:−\mathrm{2sin}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right).\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{h}\left(\frac{\mathrm{2}{x}+{h}−\mathrm{4}}{\mathrm{2}}\right)}{{h}} \\ $$$$=−\mathrm{2sin}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)\left(\frac{\mathrm{2}{x}−\mathrm{4}}{\mathrm{2}}\right) \\ $$$$=−\mathrm{2}\left({x}−\mathrm{2}\right)\mathrm{sin}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right) \\ $$$$=\left(\mathrm{4}−\mathrm{2}{x}\right)\mathrm{sin}\:\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right) \\ $$

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