derivative of 2sqrt(\sqrt{1+x^2)+1}

解

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

\frac{x}{1 + x ^{2} + 1 x ^{2} + 1}

解答ステップ

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

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

連鎖法則を適用:

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

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

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

= 2 \cdot \frac{1}{2 1 + x ^{2} + 1} \cdot \frac{x}{x ^{2} + 1}

簡素化

2 \cdot \frac{1}{2 1 + x ^{2} + 1} \cdot \frac{x}{x ^{2} + 1} : \frac{x}{1 + x ^{2} + 1 x ^{2} + 1}

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