Question

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

Answer 1

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

Answer 2

Спасибо