Question Number 187695 by mnjuly1970 last updated on 20/Feb/23
$$ \\ $$$$\:\:\:{If}\:,\:\:{f}\left(\:{x}\:\right)=\:\frac{\:\:{si}\underset{} {{n}}\:\left({cos}\:\left({x}\right)\:\right)}{\overset{} {\:}\sqrt{\:\frac{\pi}{{x}}}} \\ $$$$\:\:\:\:\Rightarrow\:\:\:{f}\:'\:\left(\frac{\:\pi}{\mathrm{2}}\:\right)\:=\:? \\ $$
Answered by horsebrand11 last updated on 20/Feb/23
$${f}\left({x}\right)=\frac{\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{\:\sqrt{\pi}}\: \\ $$$$\:{f}\:'\left(\frac{\pi}{\mathrm{2}}\right)=\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left(\frac{\pi}{\mathrm{2}}\right)}{{x}−\frac{\pi}{\mathrm{2}}} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)−\sqrt{\frac{\pi}{\mathrm{2}}}\:\mathrm{sin}\:\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\right)}{{x}−\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:.\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{2}}}\:.\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{{x}−\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(−\mathrm{sin}\:{x}\right)}{{x}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:{x}}{{x}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$
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)](https://www.tinkutara.com/question/Q187698.png)
$${f}\left({x}\right)=\frac{\sqrt{{x}}\:\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{\:\sqrt{\pi}} \\ $$$${f}\:'\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\left[\frac{\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)}{\mathrm{2}\sqrt{{x}}}\:−\sqrt{{x}}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\right] \\ $$$${f}\:'\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\left[\mathrm{0}−\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{2}}}\:.\mathrm{1}\:\right]\:=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$