(\partial)/(\partial y)(ln(2xy+4))

解答

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

\frac{x}{x y + 2}

求解步骤

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

将

x

视为常数

使用连锁律:

\frac{1}{2 x y + 4} \frac{\partial}{\partial y} (2 x y + 4)

= \frac{1}{2 x y + 4} \frac{\partial}{\partial y} (2 x y + 4)

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

= \frac{1}{2 x y + 4} \cdot 2 x

化简

\frac{1}{2 x y + 4} \cdot 2 x : \frac{x}{x y + 2}

= \frac{x}{x y + 2}

d er i v a t i v e f (x) = x^{c o s (x)}

t an g e n t f (x) = x x, at x = 18

x \to 2 + lim (\frac{1 - x}{x - 2})

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

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