derivative of x^aln(x)

解

\frac{d}{d x} (x^{a} ln (x))

a x^{a - 1} ln (x) + x^{a - 1}

解答ステップ

\frac{d}{d x} (x^{a} ln (x))

積の法則を適用:

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

f = x^{a}, g = ln (x)

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

\frac{d}{d x} (x^{a}) = a x^{a - 1}

\frac{d}{d x} (ln (x)) = \frac{1}{x}

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

\frac{1}{x} x^{a} = x^{a - 1}

= a x^{a - 1} ln (x) + x^{a - 1}