if x^y = e^y-x, then dy/dx
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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
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