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)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}}

x \to 2 lim ((x - 2)^{2})

y = cos^{3} (4 x^{6} + 10)

x^{2} - 7 x + 12

\int \frac{2 x ^{2} - x - 20}{x ^{2} + x - 6} d x

\int 8 x sec (x) tan (x) d x

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)}