derivative g(z)=10tan(z)-2cot(z)

解答

导数

g (z) = 10 tan (z) - 2 cot (z)

10 sec^{2} (z) + 2 csc^{2} (z)

求解步骤

\frac{d}{d z} (10 tan (z) - 2 cot (z))

使用微分加减法定则:

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

= \frac{d}{d z} (10 tan (z)) - \frac{d}{d z} (2 cot (z))

\frac{d}{d z} (10 tan (z)) = 10 sec^{2} (z)

\frac{d}{d z} (2 cot (z)) = - 2 csc^{2} (z)

= 10 sec^{2} (z) - (- 2 csc^{2} (z))

化简

= 10 sec^{2} (z) + 2 csc^{2} (z)

d er i v a t i v e \frac{3}{( 2 x ^{2} )}

d er i v a t i v e \frac{5}{( x ^{2} )}

d er i v a t i v e \frac{1}{1 - p}

d er i v a t i v e (1 - a)^{n}

s im pl i f y \frac{d}{dd} (n (1 - d \cdot n))