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

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

 

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