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Question Number 104350 by bemath last updated on 21/Jul/20
lim_(△x→0)  ((sin ((α+△x)^n )−sin (α^n ))/(cos ((α+△x)^n )sin (α+△x)−cos (α^n )sin (α)))
$$\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\left(\alpha+\bigtriangleup{x}\right)^{{n}} \right)−\mathrm{sin}\:\left(\alpha^{{n}} \right)}{\mathrm{cos}\:\left(\left(\alpha+\bigtriangleup{x}\right)^{{n}} \right)\mathrm{sin}\:\left(\alpha+\bigtriangleup{x}\right)−\mathrm{cos}\:\left(\alpha^{{n}} \right)\mathrm{sin}\:\left(\alpha\right)} \\ $$
Answered by john santu last updated on 21/Jul/20
lim_(△x→0)  ((sin ((α+△x)^n )−sin (α^n ))/(△x))  . lim_(△x→0)  ((△x)/(cos ((α+△x)^n sin (α+△x)−cos (α^n )sin (α)))=  (1) f(α)=sin α^n  ⇒f ′(α)=nα^(n−1) cos (α^n )  (2) g(α)=cos (α^n )sin (α)  → g ′(α)=−nα^(n−1) sin (α^n )sin(α)+  cos (α^n )cos (α)  Therefore the result is   ((nα^(n−1) cos (α^n ))/(−nα^(n−1) sin (α^n )sin (α)+cos (α^n )cos (α)))  (JS ⊛)
$$\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\left(\alpha+\bigtriangleup{x}\right)^{{n}} \right)−\mathrm{sin}\:\left(\alpha^{{n}} \right)}{\bigtriangleup{x}} \\ $$$$.\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\bigtriangleup{x}}{\mathrm{cos}\:\left(\left(\alpha+\bigtriangleup{x}\right)^{{n}} \mathrm{sin}\:\left(\alpha+\bigtriangleup{x}\right)−\mathrm{cos}\:\left(\alpha^{{n}} \right)\mathrm{sin}\:\left(\alpha\right)\right.}= \\ $$$$\left(\mathrm{1}\right)\:{f}\left(\alpha\right)=\mathrm{sin}\:\alpha^{{n}} \:\Rightarrow{f}\:'\left(\alpha\right)={n}\alpha^{{n}−\mathrm{1}} \mathrm{cos}\:\left(\alpha^{{n}} \right) \\ $$$$\left(\mathrm{2}\right)\:{g}\left(\alpha\right)=\mathrm{cos}\:\left(\alpha^{{n}} \right)\mathrm{sin}\:\left(\alpha\right) \\ $$$$\rightarrow\:{g}\:'\left(\alpha\right)=−{n}\alpha^{{n}−\mathrm{1}} \mathrm{sin}\:\left(\alpha^{{n}} \right)\mathrm{sin}\left(\alpha\right)+ \\ $$$$\mathrm{cos}\:\left(\alpha^{{n}} \right)\mathrm{cos}\:\left(\alpha\right) \\ $$$${Therefore}\:{the}\:{result}\:{is}\: \\ $$$$\frac{{n}\alpha^{{n}−\mathrm{1}} \mathrm{cos}\:\left(\alpha^{{n}} \right)}{−{n}\alpha^{{n}−\mathrm{1}} \mathrm{sin}\:\left(\alpha^{{n}} \right)\mathrm{sin}\:\left(\alpha\right)+\mathrm{cos}\:\left(\alpha^{{n}} \right)\mathrm{cos}\:\left(\alpha\right)} \\ $$$$\left({JS}\:\circledast\right) \\ $$

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