if yln(x) + xe^y = 1 dx/dt = 5 fin dy/dt when (x,y) = (1,0)

Answer 1

If you differentiate with respect to t, we get

y'ln(x) + (y/x)x' + x'ey + xeyy' substituting we get y' (0) + (0/1)(5) + 5(1) + (1)(1)y'

So,

y' = dy/dt = -5

Answer 2