derivative of x(x-aln(x+1+b^2))

解答

\frac{d}{d x} (x (x - a) ln (x + 1 + b^{2}))

2 x ln (x + b^{2} + 1) - a ln (x + b^{2} + 1) + \frac{x ( x - a )}{x + b ^{2} + 1}

求解步骤

\frac{d}{d x} (x (x - a) ln (x + 1 + b^{2}))

使用微分乘法定则:

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

f = x, g = (x - a) ln (x + 1 + b^{2})

= \frac{d x}{d x} (x - a) ln (x + 1 + b^{2}) + \frac{d}{d x} ((x - a) ln (x + 1 + b^{2})) x

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

\frac{d}{d x} ((x - a) ln (x + 1 + b^{2})) = ln (x + 1 + b^{2}) + \frac{x - a}{x + 1 + b ^{2}}

= 1 \cdot (x - a) ln (x + 1 + b^{2}) + (ln (x + 1 + b^{2}) + \frac{x - a}{x + 1 + b ^{2}}) x

化简

1 \cdot (x - a) ln (x + 1 + b^{2}) + (ln (x + 1 + b^{2}) + \frac{x - a}{x + 1 + b ^{2}}) x : 2 x ln (x + b^{2} + 1) - a ln (x + b^{2} + 1) + \frac{x ( x - a )}{x + b ^{2} + 1}

= 2 x ln (x + b^{2} + 1) - a ln (x + b^{2} + 1) + \frac{x ( x - a )}{x + b ^{2} + 1}

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

\int \frac{x ^{n}}{n !} d x

\frac{d y}{d t} - \frac{y}{2 t} = 3 t

x \to 0 lim (ln (x^{2}) - x^{- 2})

\frac{d}{d x} ((x^{3} - 9 x)^{- \frac{1}{3}})