高等数学第二章的题目，26题，求详细过程。
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26.
x = 2t-1, dx/dt = 2;
te^y+y+1 = 0, 两边对 t 求导，e^y + te^ydy/dt + dy/dt = 0,
dy/dt = -e^y/(1+te^y), 得 dy/dx = (-1/2)e^y/(1+te^y)
t = 0 时， x = -1， y = -1，dy/dt = -1/e.
d^2y/dx^2 = d(dy/dx)/dx = [d(dy/dx)/dt]/(dx/dt)
= (-1/4)d[e^y/(1+te^y)]/dt
= (-1/4){[e^y(dy/dt)(1+te^y) - e^y(e^y+te^ydy/dt)]/(1+te^y)^2}
t = 0 时，d^2y/dx^2 = (-1/4)[e^(-1)(-1/e) - e^(-2)] = 1/(2e^2)
x = 2t-1, dx/dt = 2;
te^y+y+1 = 0, 两边对 t 求导，e^y + te^ydy/dt + dy/dt = 0,
dy/dt = -e^y/(1+te^y), 得 dy/dx = (-1/2)e^y/(1+te^y)
t = 0 时， x = -1， y = -1，dy/dt = -1/e.
d^2y/dx^2 = d(dy/dx)/dx = [d(dy/dx)/dt]/(dx/dt)
= (-1/4)d[e^y/(1+te^y)]/dt
= (-1/4){[e^y(dy/dt)(1+te^y) - e^y(e^y+te^ydy/dt)]/(1+te^y)^2}
t = 0 时，d^2y/dx^2 = (-1/4)[e^(-1)(-1/e) - e^(-2)] = 1/(2e^2)
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