derivative of (ln(x+1)^{2-e^{x^2}})

解答

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

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

求解步骤

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

使用指数法则:

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

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

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

使用连锁律:

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

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

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

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

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

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

\int_{1}^{2} \frac{x}{2} - \frac{2}{x} d x

x \to \frac{π}{3} lim (tan (3 x))

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

x \to 1 lim (3 x - 1)

\frac{\partial}{\partial x} (x^{3} + z x + 10)