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

解答

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

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

求解步骤

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

将

x

视为常数

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

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

使用连锁律:

\frac{1}{y + 1} \frac{\partial}{\partial y} (y + 1)

= \frac{1}{y + 1} \frac{\partial}{\partial y} (y + 1)

\frac{\partial}{\partial y} (y + 1) = 1

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

化简

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

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

s l o p e (- 1, - 2), (0, 1)

f (t) = 4 t^{3} cos (t)

\frac{d}{d x} (3 tan (x))

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

d er i v a t i v e \frac{25 - x ^{2}}{8 x}