(\partial)/(\partial y)(-2sec^2(3x-2y+1))

解答

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

求解步骤

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

将

x

视为常数

将常数提出:

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

= - 2 \frac{\partial}{\partial y} (sec^{2} (3 x - 2 y + 1))

使用连锁律:

2 sec (3 x - 2 y + 1) \frac{\partial}{\partial y} (sec (3 x - 2 y + 1))

= 2 sec (3 x - 2 y + 1) \frac{\partial}{\partial y} (sec (3 x - 2 y + 1))

\frac{\partial}{\partial y} (sec (3 x - 2 y + 1)) = - 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= - 2 \cdot 2 sec (3 x - 2 y + 1) (- 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1))

化简

- 2 \cdot 2 sec (3 x - 2 y + 1) (- 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1)) : 8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= 8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

n = 0 \sum \infty \frac{( 8 ^{n} )}{n !}

y^{^{'}} + y = y^{2}

\frac{d}{d x} (30 x^{4} y^{7} - 24 x^{12} y^{3})

\frac{d}{d x} (- e^{x^{2}})

\frac{\partial}{\partial y} (x + ln (y))