Question

if x^y =y^x then dy by dx is
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Answer 1

Answer:

here is the explanation to the given question

Answer 2

x^y=y^x

Taking logarithms on both side

We get log (x^y)=log (y^x)

Or y.log x=x.log y

Differentiating w.r.t.x we get

y.d/dx(log x) + log x.d/dx(y) = xd/dx(log y) + logy.d/dx(x)

Or y.1/x + log x.dy/dx = x.1/y.dy/dx + log y. 1

Or dy/dx{log x - x/y } = log y - y/x

Or dy/dx = ( log y- y/x )/( log x - x/y ) = y[x.logy - y]/ x[y.log x - x]

Hope it helps :p