derivative of 2+3(x-1+3(x-1)^2+d(x-1)^3)

解答

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

3 x^{2} d + 6 x - 6 x d + 3 d - 3

求解步骤

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

化简

2 + 3 (x - 1) + 3 (x - 1)^{2} + d (x - 1)^{3} : x^{3} d + 3 x^{2} - 3 x^{2} d + 3 x d - 3 x + 2 - d

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

将

d

视为常数

使用微分加减法定则:

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

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

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

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

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

\frac{d}{d x} (3 x d) = 3 d

\frac{d}{d x} (3 x) = 3

\frac{d}{d x} (2) = 0

\frac{dd}{d x} = 0

= 3 d x^{2} + 6 x - 6 d x + 3 d - 3 + 0 - 0

化简

= 3 x^{2} d + 6 x - 6 x d + 3 d - 3

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

\frac{d}{d x} (\frac{1}{lo g _{4} ( x )})

\int_{0}^{x} sin (t^{2}) d t

\frac{d}{d x} (- 2 x^{2} - \frac{1}{2} x + 10)

y^{^{'}} + (1 + \frac{1}{t}) y = 1