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

解答

\frac{d}{d x} (lo g_{e} (x^{2} + x + 1))

\frac{2 x + 1}{x ^{2} + x + 1}

求解步骤

\frac{d}{d x} (lo g_{e} (x^{2} + x + 1))

化简

lo g_{e} (x^{2} + x + 1) : ln (x^{2} + x + 1)

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

使用连锁律:

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

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

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

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

化简

\frac{1}{x ^{2} + x + 1} (2 x + 1) : \frac{2 x + 1}{x ^{2} + x + 1}

= \frac{2 x + 1}{x ^{2} + x + 1}

\frac{d}{d x} ((- 3 x^{5} - 3)^{3})

\int x (1 - x)^{\frac{2}{3}} d x

x \to 0 + lim (2 x^{3 x})

\frac{d}{d x} (\frac{1}{x ^{3} + 4 x ^{2}})

\frac{d}{d x} (x^{2} cos (x) - 2 x sin (x) - 2 cos (x))