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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

implicit derivative

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

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

Solution steps

y^{x} = x^{y}

Treat

y

y (x)

Differentiate both sides:

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

Isolate

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

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

\frac{d y}{d t} - y = 1, y (0) = 0

t y^{^{'}} + y = 1 + y + y^{2}

\frac{d}{d x} (f (x) + g (x))

\frac{d}{d x} (x^{m})

y^{^{''}} + 8 y^{^{'}} + 15 y = 0

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