What is implicit (dy)/(dx),x^y=y^x ?

The answer to implicit (dy)/(dx),x^y=y^x is (dy)/(dx)=(y^xln(y)-yx^{y-1})/(x^yln(x)-xy^{-1+x)}

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implicit (dy)/(dx),x^y=y^x

implicit derivative

\frac{d y}{d x}, x^{y} = y^{x}

\frac{d y}{d x} = \frac{y ^{x} ln ( y ) - y x ^{y - 1}}{x ^{y} ln ( x ) - x y ^{- 1 + x}}

Solution steps

x^{y} = y^{x}

Treat

y

y (x)

Differentiate both sides:

x^{y} (ln (x) \frac{d y}{d x} + \frac{y}{x}) = (ln (y) + \frac{x \frac{d y}{d x}}{y}) y^{x}

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{y ^{x} ln ( y ) - y x ^{y - 1}}{x ^{y} ln ( x ) - x y ^{- 1 + x}}

\frac{d y}{d x} = \frac{y ^{x} ln ( y ) - y x ^{y - 1}}{x ^{y} ln ( x ) - x y ^{- 1 + x}}

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What is implicit (dy)/(dx),x^y=y^x ?
The answer to implicit (dy)/(dx),x^y=y^x is (dy)/(dx)=(y^xln(y)-yx^{y-1})/(x^yln(x)-xy^{-1+x)}