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Question Number 187695 by mnjuly1970 last updated on 20/Feb/23

     If ,  f( x )= ((  sin_  (cos (x) ))/( ^ (√( (π/x)))))      ⇒   f ′ ((( π)/2) ) = ?

$$ \\ $$$$\:\:\:{If}\:,\:\:{f}\left(\:{x}\:\right)=\:\frac{\:\:{si}\underset{} {{n}}\:\left({cos}\:\left({x}\right)\:\right)}{\overset{} {\:}\sqrt{\:\frac{\pi}{{x}}}} \\ $$$$\:\:\:\:\Rightarrow\:\:\:{f}\:'\:\left(\frac{\:\pi}{\mathrm{2}}\:\right)\:=\:? \\ $$

Answered by horsebrand11 last updated on 20/Feb/23

f(x)=(((√x) sin (cos x))/( (√π)))    f ′((π/2))=lim_(x→(π/2)) ((f(x)−f((π/2)))/(x−(π/2)))   = (1/( (√π))) lim_(x→(π/2))  (((√x) sin (cos x)−(√(π/2)) sin (cos (π/2)))/(x−(π/2)))  =(1/( (√π))) .((√π)/( (√2))) .lim_(x→(π/2))  ((sin (cos x))/(x−(π/2)))  =(1/( (√2))).lim_(x→0)  ((sin (−sin x))/x)  =(1/( (√2))) .lim_(x→0)  ((−sin x)/x)=−(1/( (√2)))

$${f}\left({x}\right)=\frac{\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{\:\sqrt{\pi}}\: \\ $$$$\:{f}\:'\left(\frac{\pi}{\mathrm{2}}\right)=\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left(\frac{\pi}{\mathrm{2}}\right)}{{x}−\frac{\pi}{\mathrm{2}}} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)−\sqrt{\frac{\pi}{\mathrm{2}}}\:\mathrm{sin}\:\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\right)}{{x}−\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:.\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{2}}}\:.\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{{x}−\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(−\mathrm{sin}\:{x}\right)}{{x}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:{x}}{{x}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$

Answered by cortano12 last updated on 20/Feb/23

f(x)=(((√x) sin (cos x))/( (√π)))  f ′(x)=(1/( (√π))) [((sin (cos x))/(2(√x))) −(√x) sin x cos (cos x)]  f ′((π/2))=(1/( (√π))) [0−((√π)/( (√2))) .1 ] =−((√2)/2)

$${f}\left({x}\right)=\frac{\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{\:\sqrt{\pi}} \\ $$$${f}\:'\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\left[\frac{\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{\mathrm{2}\sqrt{{x}}}\:−\sqrt{{x}}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\right] \\ $$$${f}\:'\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\left[\mathrm{0}−\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{2}}}\:.\mathrm{1}\:\right]\:=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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