derivative of 9/(sin(x+cos(x)))

解

\frac{d}{d x} (\frac{9}{sin ( x ) + cos ( x )})

- \frac{9 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

解答ステップ

\frac{d}{d x} (\frac{9}{sin ( x ) + cos ( x )})

定数を除去:

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

= 9 \frac{d}{d x} (\frac{1}{sin ( x ) + cos ( x )})

指数の規則を適用する:

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

= 9 \frac{d}{d x} ((sin (x) + cos (x))^{- 1})

連鎖法則を適用:

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

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

\frac{d}{d x} (sin (x) + cos (x)) = cos (x) - sin (x)

= 9 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x)))

簡素化

9 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x))) : - \frac{9 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

= - \frac{9 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}