derivative f(x)=sqrt(x^2-2x+1)

解

導関数

f (x) = x^{2} - 2 x + 1

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

解答ステップ

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

連鎖法則を適用:

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

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

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

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

簡素化

\frac{1}{2 x ^{2} - 2 x + 1} (2 x - 2) : \frac{x - 1}{x ^{2} - 2 x + 1}

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