derivative f(x)=(x+1)(2x-1)

解答

导数

f (x) = (x + 1) (2 x - 1)

4 x + 1

求解步骤

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

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x + 1, g = 2 x - 1

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

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

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

= 1 \cdot (2 x - 1) + 2 (x + 1)

化简

1 \cdot (2 x - 1) + 2 (x + 1) : 4 x + 1

= 4 x + 1

\int_{0}^{\frac{π}{8}} sin^{5} (4 x) d x

y^{^{'''}} + y^{^{''}} + 3 y^{^{'}} - 5 y = 0

\int \frac{sin ( 2 x )}{2 sin ( x )} d x

\int_{4}^{\infty} x e^{- 3 x} d x

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