Question

Пользуясь правилами и формулами дифференцирования найдите производную функции а)y=x^5/5-2/3x^3+x-7 б) y=корень из x-tgx/2+x^2cos2x в) y=(1+sinx)/(1-cosx)

Answer 1

А) 
$y' = ( \frac{1}{5} x^5- \frac{2}{3}x^3+x-7)' = ( \frac{1}{5} x^5)' - (\frac{2}{3}x^3)' + x'-7' = \\ \\ = \frac{1}{5}*5 x^4 - \frac{2}{3}*3*x^2 + 1 - 0 = \\ \\ = x^4 - 2x^2 + 1$

б)
 $y' = (\sqrt{x} - \frac{tgx}{2} + x^2cos2x)' = ({x}^{1/2})' - (\frac{tgx}{2})' + (x^2cos2x)' = \\ \\ = \frac{1}{2} {x}^{-1/2} - \frac{1}{2cos^2x} + (x^2)'*cos2x + x^{2} *(cos2x)' = \\ \\ = \frac{1}{2 \sqrt{x}} - \frac{1}{2cos^2x} + 2x*cos2x + x^{2} *(-sin2x) * 2 = \\ \\ = \frac{1}{2 \sqrt{x}} - \frac{1}{2cos^2x} + 2x*cos2x - 2x^{2} *sin2x$

в)
$y' = (\frac{(1+sinx)}{(1-cosx)})' = \frac{(1+sinx)'*(1-cosx) - (1+sinx)*(1-cosx)'}{(1-cosx)^2} = \\ \\ = \frac{cosx*(1-cosx) - (1+sinx)*sinx}{(1-cosx)^2} = \\ \\ = \frac{cosx-cos^2x - sinx -sin^2x}{(1-cosx)^2} = \frac{cosx- sinx -1}{(1-cosx)^2}$