Question

If x^y=e^x-y then dy/dx is

Answer 1

Answer:

taking log both sides

logx^y=loge^x-y

ylogx=x-yloge

differentiate both sides w.r.t x

1/y dy/DX=1-dy/DX (loge=1)

1/y dy/DX +dy/DX =1

dy/DX(1/y+1)=1

dy/DX(1+y/y)=1

dy/DX=y/1+y answer