derivative 1/(sin(x)cos(x+b+c))

解

導関数

\frac{1}{sin ( x ) cos ( x + b + c )}

- \frac{cos ( 2 x + b + c )}{cos ^{2} ( x + b + c ) sin ^{2} ( x )}

解答ステップ

\frac{d}{d x} (\frac{1}{sin ( x ) cos ( x + b + c )})

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} (sin (x) cos (x + b + c)) = cos (2 x + b + c)

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

簡素化

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

= - \frac{cos ( 2 x + b + c )}{cos ^{2} ( x + b + c ) sin ^{2} ( x )}