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

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If x^y =e^x-y than dy/dx is
  • 3 answers

Subham Mohapatra 2 years, 6 months ago

x^y=e^x-y Taking log both side ylogx = x-yloge (loge=1) ylogx = x-y Differentiate both side w.r.t x y/x + logx dy/dx = 1 - dy/dx dy/dx(logx+1)= 1-y/x dy/dx = x-y/x(logx+1) = ylogx/x(logx+1) = logx/(logx+1)^2 ANSWER

Pallavi Prasad 2 years, 6 months ago

x^y=e^x-y Taking log both side ylogx = x-yloge (loge=1) ylogx = x-y Differentiate both side w.r.t x y/x + logx dy/dx = 1 - dy/dx dy/dx(logx+1)= 1-y/x dy/dx = x-y/x(logx+1) = ylogx/x(logx+1) = logx/(logx+1)^2 ANSWER

Malvika Rai 2 years, 6 months ago

Given : x^y=e^x-y taking log both side we get, ylogx =(x-y)loge=(x-y). .....1 when x=1 then y(log1)=(1-y)=y=1 Differentiate both side w.r.t 'x' y(1/x)+logx dy/dx = 1-dy/dx =dy/dx ( logx +1) = 1 - y/x = dy/dx = x-y/x(logx +1) when x=1 then, dy/dx= 1-1/1(log1+1) = 0
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