Question

Вычислите производные функций 

 

дорого,срочно,пожалуиста 

 

 

Answer 1

а)

$f'(x) = frac{1} {2} sqrt{frac{x-1} {x^2 - 9} } (frac{x^2 - 9} {x - 1})' = frac{1} {2} sqrt{frac{x-1} {x^2 - 9} } (frac {2x(x - 1) - (x^2 - 9)} {(x-1)^2}) =$

 

$= frac{1} {2} sqrt{frac{x-1} {x^2 - 9} } (frac {x^2 -2x + 9} {(x-1)^2})$

 

б) $f'(x) = 4cos(4x + frac{pi} {4}) - frac{1} {x2}$

Answer 2

$f'(x)=(sqrt{frac{x^2-9}{x-1}})'=frac{1}{2sqrt{frac{x^2-9}{x-1}}}*(frac{x^2-9}{x-1})'=\\ frac{1}{2sqrt{frac{x^2-9}{x-1}}}*frac{(x^2-9)'*(x-1)-(x^2-9)*(x-1)'}{(x-1)^2}=\\ frac{1}{2sqrt{frac{x^2-9}{x-1}}}*frac{2x*(x-1)-(x^2-9)*1}{(x-1)^2}=\\ frac{1}{2sqrt{frac{x^2-9}{x-1}}}*frac{2x^2-2x-x^2+9}{(x-1)^2}=\\ frac{1}{2sqrt{frac{x^2-9}{x-1}}}*frac{x^2-2x+9}{(x-1)^2}=\\ frac{x^2-2x+9}{2sqrt{frac{x^2-9}{x-1}}*(x-1)^2}}$

 

$f'(x)=(sin(4x+frac{pi}{4})+frac{1}{x})'=(sin(4x+frac{pi}{4}))'+(frac{1}{x})'=\\ cos(4x+frac{pi}{4})*(4x+frac{pi}{4})'-frac{1}{x^2}=\\ 4cos(4x+frac{pi}{4})-frac{1}{x^2}$