derivative of 4x^2+2x+1n(x-3x^2-7x-3)

解答

\frac{d}{d x} (4 x^{2} + 2 x + 1 n (x) - 3 x^{2} - 7 x - 3)

2 x + n - 5

求解步骤

\frac{d}{d x} (4 x^{2} + 2 x + 1 \cdot n (x) - 3 x^{2} - 7 x - 3)

化简

4 x^{2} + 2 x + 1 \cdot n (x) - 3 x^{2} - 7 x - 3 : x^{2} + n x - 5 x - 3

= \frac{d}{d x} (x^{2} + n x - 5 x - 3)

使用微分加减法定则:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (n x) - \frac{d}{d x} (5 x) - \frac{d}{d x} (3)

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

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

\frac{d}{d x} (5 x) = 5

\frac{d}{d x} (3) = 0

= 2 x + n - 5 - 0

化简

= 2 x + n - 5

d er i v a t i v e y = 4 x^{3} (2 - 3 x)

d er i v a t i v e y = \frac{t ^{3}}{3}

\frac{d}{d n} (i^{2 n + 3})

d er i v a t i v e y = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + \frac{x}{4}

d er i v a t i v e y = ln ((x + 3)^{2})