derivative of 3x^2+2x+1

解答

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

6 x + 2

求解步骤

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

使用微分加减法定则:

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

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

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (2 x) = 2

\frac{d}{d x} (1) = 0

= 6 x + 2 + 0

化简

= 6 x + 2

t^{2} y^{^{'}} + t y = 5

l a pl a ce t r an s f or m x sin (3 x)

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

t a y l or x \cdot ln (x)

\int \frac{x + 1}{x ^{3}} d x