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

解

\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}

解答ステップ

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

積の法則を適用:

(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

簡素化

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}