(\partial)/(\partial x)(x^2tan(4x^4y))

解答

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

2 x tan (4 x^{4} y) + 16 x^{5} y sec^{2} (4 x^{4} y)

求解步骤

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

将

y

视为常数

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = tan (4 x^{4} y)

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (tan (4 x^{4} y)) = sec^{2} (4 x^{4} y) \cdot 16 y x^{3}

= 2 x tan (4 x^{4} y) + sec^{2} (4 x^{4} y) \cdot 16 y x^{3} x^{2}

sec^{2} (4 x^{4} y) \cdot 16 y x^{3} x^{2} = 16 x^{5} y sec^{2} (4 x^{4} y)

= 2 x tan (4 x^{4} y) + 16 x^{5} y sec^{2} (4 x^{4} y)

\int \frac{1}{x ^{2} x ^{2} - 64} d x

\int \frac{1}{t ^{2} - 8 t + 25} d t

\frac{d y}{d x} = \frac{2 x}{y ^{2}}

x \to 2 lim (6 \cdot a + x)

in v erse l a pl a ce \frac{( 18 )}{s ( s ^{2} + 6 s + 18 )}