derivative of 9x+1/x-4

解答

\frac{d}{d x} (9 x + \frac{1}{x} - 4)

- \frac{1}{x ^{2}} + 9

求解步骤

\frac{d}{d x} (9 x + \frac{1}{x} - 4)

使用微分加减法定则:

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

= \frac{d}{d x} (9 x) + \frac{d}{d x} (\frac{1}{x}) - \frac{d}{d x} (4)

\frac{d}{d x} (9 x) = 9

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

\frac{d}{d x} (4) = 0

= 9 - \frac{1}{x ^{2}} - 0

化简

= - \frac{1}{x ^{2}} + 9

x \to 3 lim (4 - x)

y = \frac{1}{( x ^{2} - 2 ) ( x ^{2} + x + 5 )}

\int \frac{7 x + 6}{x ^{2} + 3 x} d x

\frac{d}{d x} (\frac{400}{1 + \frac{1}{8} x} + 200)

d er i v a t i v e \frac{, at}{B t ^{2} + C t ^{3}}