derivative of |x|^{|x|}

解答

\frac{d}{d x} (∣ x ∣^{∣ x ∣})

x ∣ x ∣^{∣ x ∣ - 1} (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^{∣ x ∣ l n ∣ x ∣}

= \frac{d}{d x} (e^{∣ x ∣ l n ∣ x ∣})

使用连锁律:

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

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

\frac{d}{d x} (∣ x ∣ ln ∣ x ∣) = \frac{x ( ln ∣ x ∣ + 1 )}{∣ x ∣}

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

化简

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

= x ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1)

\frac{\partial}{\partial x} (\frac{8}{x ^{2}})

d er i v a t i v e f (x) = 3 x^{2} + x + 5

d er i v a t i v e f (x) = - 5 x^{2} + 2 x - 6

x \to 0 + lim (\frac{∣ x - 0 ∣}{0 - x})

\int_{2}^{\infty} x e^{- 4 x} d x