derivative of 6+cos(sqrt(3x))

解答

\frac{d}{d x} (6 + cos (3 x))

- \frac{3 sin ( 3 x )}{2 x}

求解步骤

\frac{d}{d x} (6 + cos (3 x))

化简

6 + cos (3 x) : 6 + cos (3 x)

= \frac{d}{d x} (6 + cos (3 x))

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (6) + \frac{d}{d x} (cos (3 x))

\frac{d}{d x} (6) = 0

\frac{d}{d x} (cos (3 x)) = - \frac{3 sin ( 3 x )}{2 x}

= 0 - \frac{3 sin ( 3 x )}{2 x}

化简

= - \frac{3 sin ( 3 x )}{2 x}

x \to 0 + lim (\frac{3 ^{x} - 1}{x ^{2}})

\int \frac{x ^{3}}{a} d x

d er i v a t i v e x 2 x^{2} + 7

\frac{\partial}{\partial x} (∣ x ∣ y + 2 x)

\int \frac{1 + x}{x} d x