derivative of \sqrt[m+n]{(1-x^m(1+x)^n})

解

\frac{d}{d x} (m + n (1 - x)^{m} (1 + x)^{n})

\frac{( - m ( 1 + x ) ^{n} ( 1 - x ) ^{m - 1} + n ( 1 - x ) ^{m} ( 1 + x ) ^{n - 1} ) ( ( 1 - x ) ^{m} ( 1 + x ) ^{n} ) ^{\frac{1 - m - n}{m + n}}}{m + n}

解答ステップ

\frac{d}{d x} (m + n (1 - x)^{m} (1 + x)^{n})

連鎖法則を適用:

\frac{( ( 1 - x ) ^{m} ( 1 + x ) ^{n} ) ^{\frac{1 - m - n}{m + n}}}{m + n} \frac{d}{d x} ((1 - x)^{m} (1 + x)^{n})

= \frac{( ( 1 - x ) ^{m} ( 1 + x ) ^{n} ) ^{\frac{1 - m - n}{m + n}}}{m + n} \frac{d}{d x} ((1 - x)^{m} (1 + x)^{n})

\frac{d}{d x} ((1 - x)^{m} (1 + x)^{n}) = - m (1 + x)^{n} (1 - x)^{m - 1} + n (1 - x)^{m} (1 + x)^{n - 1}

= \frac{( ( 1 - x ) ^{m} ( 1 + x ) ^{n} ) ^{\frac{1 - m - n}{m + n}}}{m + n} (- m (1 + x)^{n} (1 - x)^{m - 1} + n (1 - x)^{m} (1 + x)^{n - 1})

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( - m ( 1 + x ) ^{n} ( 1 - x ) ^{m - 1} + n ( 1 - x ) ^{m} ( 1 + x ) ^{n - 1} ) ( ( 1 - x ) ^{m} ( 1 + x ) ^{n} ) ^{\frac{1 - m - n}{m + n}}}{m + n}