Question

помогите решить производные
$1)y(x)= frac{sinx}{cosx} \ 2)y(x)=sin( frac{5 pi }{3} x) \ 3)y(x)= sqrt{x} - frac{3}{x}+ frac{9}{x^2} \ 4)y(x)= frac{x^2+4}{x^2-4} \ 5)f(x)= frac{ sqrt{x} }{ sqrt{x} +1}$

Answer 1

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Answer 2

$y= cfrac{sin x}{cos x} \ y'= cfrac{(sin x)'cdotcos x-sin xcdot(cos x)'}{cos^2x} = \ = cfrac{cos xcdotcos x-sin xcdot(-sin x)}{cos^2x} = cfrac{cos^2 x+sin ^2x}{cos^2x} = frac{1}{cos^2x}$

$y=sin( frac{ 5pi }{3} x) \ y'=cos( frac{ 5pi }{3} x)cdot( frac{ 5pi }{3} x)'= frac{ 5pi }{3}cos( frac{ 5pi }{3} x)$

$y= sqrt{x} - frac{3}{x}+ frac{9}{x^2}=x^ frac{1}{2} - 3x^{-1}+9x^{-2} \ y'= frac{1}{2}x^ {frac{1}{2}-1}-(- 3x^{-1-1})+9cdot(-2x^{-2-1})= \ =frac{1}{2}x^ {-frac{1}{2}}+ 3x^{-2}-18x^{-3}= frac{1}{2 sqrt{x} } + frac{3}{x^2} - frac{18}{x^3}$

$y= cfrac{x^2+4}{x^2-4} \ y= cfrac{(x^2+4)'(x^2-4)-(x^2+4)(x^2-4)'}{(x^2-4)^2}= cfrac{2x(x^2-4)-2x(x^2+4)}{(x^2-4)^2}= \ =cfrac{2x(x^2-4-x^2-4)}{(x^2-4)^2}=cfrac{2xcdot(-8)}{(x^2-4)^2}=- cfrac{16x}{(x^2-4)^2}$

$y= cfrac{ sqrt{x} }{ sqrt{x} +1} \ y'= cfrac{( sqrt{x} )'(sqrt{x} +1)- sqrt{x} (sqrt{x} +1)'}{( sqrt{x} +1)^2} = cfrac{ frac{1}{2 sqrt{x} }(sqrt{x} +1)- sqrt{x} cdot frac{1}{2 sqrt{x} }}{( sqrt{x} +1)^2} = \ = cfrac{ frac{1}{2 sqrt{x} }(sqrt{x} +1- sqrt{x} ) }{( sqrt{x} +1)^2} = cfrac{ frac{1}{2 sqrt{x} } }{( sqrt{x} +1)^2} = cfrac{1}{2 sqrt{x}( sqrt{x} +1)^2}$