derivative y=5ln(4x^2+1)

解

導関数

y = 5 ln (4 x^{2} + 1)

\frac{40 x}{4 x ^{2} + 1}

解答ステップ

\frac{d}{d x} (5 ln (4 x^{2} + 1))

定数を除去:

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

= 5 \frac{d}{d x} (ln (4 x^{2} + 1))

連鎖法則を適用:

\frac{1}{4 x ^{2} + 1} \frac{d}{d x} (4 x^{2} + 1)

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

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= 5 \cdot \frac{1}{4 x ^{2} + 1} \cdot 8 x

簡素化

5 \cdot \frac{1}{4 x ^{2} + 1} \cdot 8 x : \frac{40 x}{4 x ^{2} + 1}

= \frac{40 x}{4 x ^{2} + 1}