derivative of x^3ln(x^4+1)

解

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

3 x^{2} ln (x^{4} + 1) + \frac{4 x ^{6}}{x ^{4} + 1}

解答ステップ

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

積の法則を適用:

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

f = x^{3}, g = ln (x^{4} + 1)

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

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

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

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

\frac{4 x ^{3}}{x ^{4} + 1} x^{3} = \frac{4 x ^{6}}{x ^{4} + 1}

= 3 x^{2} ln (x^{4} + 1) + \frac{4 x ^{6}}{x ^{4} + 1}