derivative of xarctan(2x-1/4 ln(1+4x^2))

解答

\frac{d}{d x} (x arctan (2 x) - \frac{1}{4} ln (1 + 4 x^{2}))

arctan (2 x)

求解步骤

\frac{d}{d x} (x arctan (2 x) - \frac{1}{4} ln (1 + 4 x^{2}))

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (x arctan (2 x)) - \frac{d}{d x} (\frac{1}{4} ln (1 + 4 x^{2}))

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

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

= arctan (2 x) + \frac{2 x}{4 x ^{2} + 1} - \frac{2 x}{1 + 4 x ^{2}}

化简

arctan (2 x) + \frac{2 x}{4 x ^{2} + 1} - \frac{2 x}{1 + 4 x ^{2}} : arctan (2 x)

= arctan (2 x)

t an g e n t 9 x^{\frac{3}{5}} - 2 x^{\frac{6}{7}}, at x = 3

t an g e n t f (x) = tan (x), (\frac{3 π}{4}, - 1)

x \to 2 π lim (csc (x))

d er i v a t i v e 9 - x^{2}

\frac{d y}{d x} = 9 y \cdot e^{x + 9}