Помогите, срочно! Нужно найти производную

1) $y' = (2x^{2})' - (\frac{1}{3} x^{3})'+(5x)'-(7)'=4x- \frac{1}{3}*3 x^{2}+5=4x-x^{2}+5$
2) $y' = ( \frac{2x-1}{5+7x})' = \frac{(2x-1)'(5+7x)-(2x-1)(5+7x)'}{(5+7x)^{2}} = \frac{2(5+7x)-7(2x-1)}{(5+7x)^{2}} =$ $\frac{10+14x-14x+7}{(5+7x)^{2}} = \frac{17}{(5+7x)^{2}}$
3) $y'=((4x-2)(3x^{2}+1))'=(4x-2)'(3x^{2}+1)+(4x-2)(3x^{2}+1)'=$ $4(3x^{2}+1)+6x(4x-2)=12x^{2}+4+24x^{2}-12x=36x^{2}-8x$
4) $y'= (\frac{ \sqrt{x} *x}{ \sqrt[3]{x} } + \sqrt{2} )'= \frac{( \sqrt{x} *x)'* \sqrt[3]{x} + \sqrt{x} *x( \sqrt[3]{x})' }{ \sqrt[3]{x^{2}} }= \frac{(x^ \frac{3}{2})'* x^ \frac{1}{3} +x^ \frac{3}{2} ( x^ \frac{1}{3} )' }{ x^ \frac{2}{3} } =$ $\frac{ \frac{3}{2}x^ \frac{1}{2}*x^ \frac{1}{3} + \frac{1}{3} x^ \frac{-2}{3} *x^ \frac{3}{2} }{x^ \frac{2}{3} } = \frac{ \frac{3}{2} x^ \frac{5}{6}+ \frac{1}{3}x^ \frac{5}{6} }{x^ \frac{2}{3} } = \frac{ \frac{11}{6} x^ \frac{5}{6} }{x^ \frac{2}{3} } = \frac{11}{6}x^ \frac{1}{6} = \frac{11}{6} \sqrt[6]{x}$
5) $y'=( e^{x}* 3^{x})' + (\frac{lnx}{1+cosx})' -(tg \frac{\pi}{4} )'=$ $( e^{x})'* 3^{x}+e^{x}* (3^{x})'+ \frac{(lnx)'(1+cosx)-lnx(1+cosx)'}{(1+cosx)^{2}} =$ $e^{x} *3^x+e^x *3^x*ln3 + \frac{ \frac{1}{x}(1+cosx) - lnx*(-sinx) }{(1+cosx)^2} =$ $e^x*3^x(1+ln3)+ \frac{1+cosx+x*lnx*sinx}{x*(1+cosx)^2}$