Question

if x^y = e^y-x, then dy/dx​

Answer 1

Answer:

1 + logx = 0 (not sure)

Step-by-step explanation:

on taking log we get

ylogx = (y-x)logx

ylogx = ylogx - xlogx

0 = xlogx

on doing differentiation,

0 = x/x + logx

0= 1 + logx