Question

Учителя Светила Прошу!!!!!!сделайте 540

Answer 1

$f(x)= frac{Cos ^{2}x }{1+Sin ^{2}x } \\f'(x)= frac{(Cos ^{2}x)'*(1+Sin ^{2}x)-Cos ^{2}x*(1+Sin ^{2} x)}{(1+Sin ^{2}x) ^{2} } =$$frac{2Cosx*(Cosx)'*(1+Sin ^{2}x)-Cos ^{2}x*2Sinx*(Sinx)' }{(1+Sin ^{2}x) ^{2} } =$ frac{-2CosxSinx(1+Sin ^{2} x)-2Cos ^{3} xSinx}{(1+Sin ^{2}x) ^{2} }[/tex] =[/tex] frac{-2CosxSinx-2CosxSin ^{3}x-2Cos ^{3} xSinx }{(1+Sin ^{2}x) ^{2} } [/tex] frac{-2CosxSinx(1+Sin ^{2}x+Cos ^{2} x) }{(1+Sin ^{2}x) ^{2} } = frac{-4CosxSinx}{(1+Sin ^{2} x) ^{2} }\\f( frac{ pi }{4}) = frac{Cos ^{2} frac{ pi }{4} }{1+Sin ^{2} frac{ pi }{4} } = frac{( frac{1}{ sqrt{2} }) ^{2} }{1+( frac{1}{ sqrt{2} }) ^{2} }= frac{ frac{1}{2} }{ frac{3}{2} }= frac{1}{3}\\f'( frac{ pi }{4}) = frac{-4*Cos frac{ pi }{4} Sin frac{ pi }{4} }{(1+Sin ^{2} frac{ pi }{4}) ^{2} } = [/tex]$frac{-4* frac{1}{ sqrt{2} }* frac{1}{ sqrt{2} } }{[1+( frac{1}{ sqrt{2} }) ^{2}] ^{2} }= frac{-2}{( frac{3}{2}) ^{2} } = frac{-2}{ frac{9}{4} }=- frac{8}{9} \\f( frac{ pi }{4})-3f'( frac{ pi }{4} ) = frac{1}{3}-3*(- frac{8}{9})= frac{1}{3}+ frac{8}{3}= frac{9}{3}=3$

Answer 2

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Answer 3

огромное вам спасибо