derivative of 3x(10x^3-6/(x+10))

解答

\frac{d}{d x} (3 x (10 x^{3} - \frac{6}{x + 10}))

\frac{3 ( 40 x ^{5} + 800 x ^{4} + 4000 x ^{3} - 60 )}{( x + 10 ) ^{2}}

求解步骤

\frac{d}{d x} (3 x (10 x^{3} - \frac{6}{x + 10}))

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (x (10 x^{3} - \frac{6}{x + 10}))

使用微分乘法定则:

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

f = x, g = 10 x^{3} - \frac{6}{x + 10}

= 3 (\frac{d x}{d x} (10 x^{3} - \frac{6}{x + 10}) + \frac{d}{d x} (10 x^{3} - \frac{6}{x + 10}) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (10 x^{3} - \frac{6}{x + 10}) = 30 x^{2} + \frac{6}{( x + 10 ) ^{2}}

= 3 (1 \cdot (10 x^{3} - \frac{6}{x + 10}) + (30 x^{2} + \frac{6}{( x + 10 ) ^{2}}) x)

化简

3 (1 \cdot (10 x^{3} - \frac{6}{x + 10}) + (30 x^{2} + \frac{6}{( x + 10 ) ^{2}}) x) : \frac{3 ( 40 x ^{5} + 800 x ^{4} + 4000 x ^{3} - 60 )}{( x + 10 ) ^{2}}

= \frac{3 ( 40 x ^{5} + 800 x ^{4} + 4000 x ^{3} - 60 )}{( x + 10 ) ^{2}}

\frac{d}{d x} (- 5 cos (5 x))

x \to - \infty lim (5^{x} + 2)

\int_{0}^{\frac{1}{2}} 3 arccos (x) d x

\frac{\partial}{\partial y} (x \cdot ln (y - z))

\int \frac{x ^{3}}{x ^{4} + 3 x ^{2} + 2} d x