derivative of x^3ln(1+x^2)

解

\frac{d}{d x} (x^{3} ln (1 + x^{2}))

3 x^{2} ln (1 + x^{2}) + \frac{2 x ^{4}}{1 + x ^{2}}

解答ステップ

\frac{d}{d x} (x^{3} ln (1 + x^{2}))

積の法則を適用:

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

f = x^{3}, g = ln (1 + x^{2})

= \frac{d}{d x} (x^{3}) ln (1 + x^{2}) + \frac{d}{d x} (ln (1 + x^{2})) x^{3}

\frac{d}{d x} (x^{3}) = 3 x^{2}

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

= 3 x^{2} ln (1 + x^{2}) + \frac{2 x}{1 + x ^{2}} x^{3}

\frac{2 x}{1 + x ^{2}} x^{3} = \frac{2 x ^{4}}{1 + x ^{2}}

= 3 x^{2} ln (1 + x^{2}) + \frac{2 x ^{4}}{1 + x ^{2}}