derivative y=\sqrt[x]{x}

解答

导数

y = x x

x^{\frac{1 - 2 x}{x}} (- ln (x) + 1)

求解步骤

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

使用指数法则:

a^{b} = e^{b l n (a)}

x x = e^{\frac{1}{x} l n (x)}

= \frac{d}{d x} (e^{\frac{1}{x} l n (x)})

使用连锁律:

e^{\frac{1}{x} l n (x)} \frac{d}{d x} (\frac{1}{x} ln (x))

= e^{\frac{1}{x} l n (x)} \frac{d}{d x} (\frac{1}{x} ln (x))

\frac{d}{d x} (\frac{1}{x} ln (x)) = \frac{- ln ( x ) + 1}{x ^{2}}

= e^{\frac{1}{x} l n (x)} \frac{- ln ( x ) + 1}{x ^{2}}

化简

e^{\frac{1}{x} l n (x)} \frac{- ln ( x ) + 1}{x ^{2}} : x^{\frac{1 - 2 x}{x}} (- ln (x) + 1)

= x^{\frac{1 - 2 x}{x}} (- ln (x) + 1)

\int \frac{1}{( x ^{2} + 64 ) ^{2}} d x

\frac{d}{d x} (e^{\frac{x}{10}})

x \to o - lim (4)

\frac{d}{d x} ((x^{2} + 1)^{6})

(x^{2} + 2 x y + y^{2}) d x + (2 x y + x^{2} - 1) d y = 0