d/(d{y)}(-9sin(3{x}+{y}{z}))

解答

\frac{d}{d y} (- 9 sin (3 x + y z))

- 9 cos (3 x + y z) z

求解步骤

\frac{d}{d y} (- 9 sin (3 x + y z))

将常数提出:

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

= - 9 \frac{d}{d y} (sin (3 x + y z))

使用连锁律:

cos (3 x + y z) \frac{d}{d y} (3 x + y z)

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

\frac{d}{d y} (3 x + y z) = z

= - 9 cos (3 x + y z) z

\frac{\partial ^{2}}{\partial x \partial y} (ln (4 + x^{2} y^{2}))

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

x \to 2 lim (7)

d er i v a t i v e (2 x - 3)^{4} (x^{2} + x + 1)^{5}

\frac{d}{d t} (A t e^{t})