(\partial)/(\partial x)(ln(x^2+1)+y^2-z)

解答

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

\frac{2 x}{x ^{2} + 1}

求解步骤

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

将

y, z

视为常数

使用微分加减法定则:

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

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

\frac{\partial}{\partial x} (ln (x^{2} + 1)) = \frac{2 x}{x ^{2} + 1}

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (z) = 0

= \frac{2 x}{x ^{2} + 1} + 0 - 0

化简

= \frac{2 x}{x ^{2} + 1}

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

a re a x^{2} - 4, y = 0

\int 64 x^{2} cos (4 x) d x

d er i v a t i v e g (t) = \frac{9}{t}

\frac{d y}{d x} + 6 x^{2} y = e^{- 2 x^{3}}