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

解答

导数

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

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

求解步骤

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

使用指数法则:

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

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

使用连锁律:

- \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)

化简

- \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 )}

\int_{0}^{1} arcsin (x) - sin (x) d x

\int (\frac{3 x ^{2} + 3 x + 1}{x}) d x

s l o p e f (x) = \frac{x ^{2}}{x + 9}

\int \frac{1}{cos ^{2} ( 7 x )} d x

(6 x^{2} + y - 1) d x - (14 y - x) d y = 0, y (1) = 0