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人気のある微分積分 >

(\partial)/(\partial z)(e^{xy}+z)

解

\frac{\partial}{\partial z} (e^{x y} + z)

解

1

解答ステップ

\frac{\partial}{\partial z} (e^{x y} + z)

x, y

を定数として扱う

和/差の法則を適用:

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

= \frac{\partial}{\partial z} (e^{x y}) + \frac{\partial}{\partial z} (z)

\frac{\partial}{\partial z} (e^{x y}) = 0

\frac{\partial}{\partial z} (z) = 1

= 0 + 1

簡素化

= 1

人気の例

derivative of (x+2^2+(10-3x)^2)

\frac{d}{d x} ((x + 2)^{2} + (10 - 3 x)^{2})

sum from n=1 to infinity of 1+1/(n+1)

n = 1 \sum \infty 1 + \frac{1}{n + 1}

limit as x approaches 0 of ((2-x)^3-8)/x

x \to 0 lim (\frac{( 2 - x ) ^{3} - 8}{x})

inverselaplace 5/(s^2+49)

in v erse l a pl a ce \frac{5}{s ^{2} + 49}

y(1+x^2)y^'-x(9+y^2)=0

y (1 + x^{2}) y^{^{'}} - x (9 + y^{2}) = 0