(\partial)/(\partial x)(z/x ln(y)+ln(x))

解答

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

- \frac{z ln ( y )}{x ^{2}} + \frac{1}{x}

求解步骤

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

化简

\frac{z}{x} ln (y) + ln (x) : \frac{z ln ( y )}{x} + ln (x)

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

将

z, y

视为常数

使用微分加减法定则:

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

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

\frac{\partial}{\partial x} (\frac{z ln ( y )}{x}) = - \frac{z ln ( y )}{x ^{2}}

\frac{\partial}{\partial x} (ln (x)) = \frac{1}{x}

= - \frac{z ln ( y )}{x ^{2}} + \frac{1}{x}

\int \frac{x}{2 + x} d x

d er i v a t i v e f (x) = (- x^{3} - x^{2} + x) cos (x)

\frac{d y}{d x} = x + \frac{1}{5} y

\frac{d}{d x} (\frac{x ^{2} - 4}{x + 1})

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = 0