Question

найти y", если x=t^2+2t y=ln(1+t)

Answer 1

y=ln(1+t+t^2+2t)=ln(t^2+3t=1);
y'=(1/t^2+3t+1))*(2t+3)= 2t+3/(t^2+3t+1);
y''=(2(t^2+3t+1)-(2t+3)(2t+3))/(t^2+3t+1)^2=
=(2t^2+6t+2-4t^2-12t-9)/(t^2+3t+1)^2 = -(2t^2+6t+7)/(t^2+3t+1)^2/