Найти производную 1-ого и 2-ого порядка.Срочно!

Решите задачу:

$1. \\ y'=e^x+4 \\ y''=e^x \\ 2. \\ \left \{ {{x=e^tcost} \atop {y=e^tsint}} \right \\ dx=e^tcost-e^tsint \\ dy=e^tsint+e^tcost \\ y'_x= \frac{dy}{dx}= \frac{e^tsint+e^tcost}{e^tcost-e^tsint}= \frac{sint+cost}{cost-sint} \\ \left \{ {{y'_x=\frac{sint+cost}{cost-sint} } \atop {x=e^tcost}} \right \\ y''_{xx}=\frac{dy'_x}{dx} =\frac{(y'_x)'_t}{x'_t} \\ (y'_x)'_t=(\frac{sint+cost}{cost-sint})'_t= \\ =\frac{(sint+cost)'(cost-sint)-(cost-sint)'(sint+cost)}{(cost-sint)^2}= \\ =\frac{(cost-sint)(cost-sint)-(-sint-cost)(sint+cost)}{cos^2t-2sintcost+sin^2t}= \\ =\frac{cos^2t-2sintcost+sin^2t+sin^2t+2costsint+cos^2t)}{1-sin2t}= \frac{2(sin^2t+cos^2t)}{1-sin2t}=\frac{2}{1-sin2t} \\ y''_{xx}=\frac{2}{(1-sin2t)(e^tcost-e^tsint)}=\frac{2}{e^t(1-sin2t)(cost-sint)}$
$\left \{ {{y''_{xx}=\frac{2}{e^t(1-sin2t)(cost-sint)} } \atop {x=e^tcost}} \right$